Advanced
Integration
Manual Title: Advanced Integration
Author: Dan Hamilton
Editor: John Hamilton
Cover design by: John Hamilton
Copyright  2002
All rights reserved.
Printed in the United States of America.
No part of this manual may be reproduced, stored in a retrieval system, or transmitted in any
form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without
the prior written permission of the author. Request for permission or further information should
be addressed to Hamilton Education Guides via info@hamiltoneducationguides.com.
First published in 2002
Library of Congress Catalog Card Number 20022-81883
Library of Congress Cataloging-in-Publication Data
ISBN 978-1-5323-9408-9
Hamilton Education Guides
Book Series
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eBook and paperback versions available in the Amazon Kindle Store
Hamilton Education Guides
Manual Series
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eManual versions available in the Amazon Kindle Store
Hamilton Education Guides
Manual Series
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eManual versions available in the Amazon Kindle Store
Contents
Advanced Integration
Quick Reference to Problems 1
1.1
Integration by Parts ................................................................................................... 2
1.2
Integration Using Trigonometric Substitution........................................................ 23
1.3
Integration by Partial Fractions ............................................................................... 35
Case I - The Denominator Has Distinct Linear Factors 35
Case II - The Denominator Has Repeated Linear Factors 42
Case III - The Denominator Has Distinct Quadratic Factors 49
Case IV - The Denominator Has Repeated Quadratic Factors 59
1.4
Integration of Hyperbolic Functions ........................................................................ 65
Appendix – Exercise Solutions.............................................................................................. 94
Section 1.1 94
Section 1.4 113
Hamilton Education Guides
Section 1.2 102
Section 1.3 106
i
Acknowledgments
The primary motivating factor in writing the Hamilton Education Guides manual series is
to provide students with specific subjects on mathematics. I am grateful to John Hamilton for his
editorial comments, cover design, and suggestions on easier presentation of the topics. I would
also like to acknowledge the original contributors of the Hamilton Education Guides books for
their editorial reviews. Finally, I would like to thank my family for their understanding and
patience in allowing me to prepare this manual.
Hamilton Education Guides
ii
Introduction and Overview
It is my belief that the key to learning mathematics is through positive motivation.
Students can be greatly motivated if subjects are presented concisely and the problems are solved
in a detailed step by step approach. This keeps students motivated and provides a great deal of
encouragement in wanting to learn the next subject or to solve the next problem. During my
teaching career, I found this method to be an effective way of teaching. I hope by presenting
equations in this format, more students will become interested in the subject of mathematics.
This manual is a chapter from my Calculus 1 – Differentiation and Integration book with
the primary focus on the subject of advanced integration. The scope of this manual is intended
for educational levels ranging from the 12th grade to adult. The manual can also be used by
students in home study programs, parents, teachers, special education programs, preparatory
schools, and adult educational programs including colleges and universities as a supplementary
manual. A fundamental understanding of basic mathematical operations such as addition,
subtraction, multiplication, and division is required.
This manual addresses advanced integrals and how they are simplified and mathematically
operated. Students learn topics that include the integration of functions using the Integration by
Parts, Trigonometric Substitution, and Partial Fractions methods; and how to find the integral of
hyperbolic functions. Detailed solutions to the exercises are provided in the Appendix. Students
are encouraged to solve each problem in the same detail and step by step format as shown in the
text.
It is my hope that all Hamilton Education Guides books and manuals stand apart in their
understandable treatment of the presented subjects and for their clarity and special attention to
detail. I hope readers of this manual will find it useful.
With best wishes,
Dan Hamilton
Hamilton Education Guides
iii
Advanced Integration
Quick Reference to Problems
Advanced Integration
Quick Reference to Problems
1.1
Integration by Parts ................................................................................................... 2
∫ e cos 2 x dx = ; ∫ e sin x dx = ; ∫ x cos 3x dx =
2x
1.2
x
Integration Using Trigonometric Substitution........................................................ 23
x 2 dx
dx
dx
∫ 36 − x 2 = ; ∫ ( 9 + x 2 )2 = ; ∫ x 4 x 2 −1 =
1.3
Integration by Partial Fractions ............................................................................... 35
Case I - The Denominator Has Distinct Linear Factors 35
∫
x +1
dx = ;
x (x − 2 )(x + 3)
1
∫ (x + 1)(x + 2)
dx = ;
∫
x 2 +1
dx =
x(x − 1)(x + 1)
Case II - The Denominator Has Repeated Linear Factors 42
1
x+3
5
∫ x (x − 1)2 dx = ; ∫ x 2 (x − 1) dx = ; ∫ x (x − 1)2 dx =
Case III - The Denominator Has Distinct Quadratic Factors 49
x2 − x + 3
1
1
∫ x (x 2 + 1) dx = ; ∫ x (x 2 + 25) dx = ; ∫ x 2 (x 2 + 16) dx =
Case IV - The Denominator Has Repeated Quadratic Factors 59
x2
∫ (x 2 + 1)2
1.4
dx = ;
x 2 +1
∫ (x 2 + 4)2
dx = ;
x3
∫ (x 2 + 2)2 dx =
Integration of Hyperbolic Functions ........................................................................ 65
1
∫ cosh 5 x dx = ; ∫ (sinh 4 x + cosh 2 x ) dx = ; ∫ x csc h x dx =
Hamilton Education Guides
2
2 3
1
Advanced Integration
The objective of this manual is to improve the student’s ability to solve additional problems
involving integration. A method used to integrate functions, known as Integration by Parts, is
addressed in Section 1.1. Integration of functions using the Trigonometric Substitution method
is discussed in Section 1.2. Integration of functions using the Partial Fractions technique is
addressed in Section 1.3. Four different cases, depending on the denominator having distinct
linear factors, repeated linear factors, distinct quadratic factors, or repeated quadratic factors, are
addressed in this section. Finally, integration of hyperbolic functions is discussed in Section 1.4.
Each section is concluded by solving examples with practice problems to further enhance the
student’s ability.
1.1
Integration by Parts
Integration by parts is a technique for replacing hard to integrate integrals by ones that are easier
to integrate. This technique applies mainly to integrals that are in the form of ∫ f (x ) g (x ) dx
where in most cases f (x ) can be differentiated several times to become zero and g (x ) can be
integrated several times without difficulty.
For example, given the integral
∫ x e dx the
2 3x
function f (x ) = x 2 can be differentiated three times to become zero and the function g (x ) = e 3 x
can be integrated several times easily. On the other hand, integrals such as ∫ e 2 x cos 2 x dx and
∫e
−3 x
sin 3 x dx do not fall under the category described above. In this section we will learn how to
apply Integration by Parts method in solving various integrals. The formula for integration by
parts comes from the product rule, i.e.,
d
( uv ) = u dv + v du
dx
dx
dx
where u and v are differentiable functions of x , multiplying both sides of the equation by dx
we obtain
rearranging the terms we then have
d ( uv ) = u dv + v du
u dv = d ( uv ) − v du
integrating both sides of the equation we obtain
∫ u dv = uv − ∫ v du
The above formula is referred to as the Integration by Parts Formula. Note that in using the
above equality we must first select dv such that it is easily integrable and second ensure that
∫ u dv is easier to evaluate than ∫ u dv . In the following examples we will solve problems using
the Integration by Parts method.
Example 1.1-1: Evaluate the following indefinite integrals:
a. ∫ x e x dx =
b. ∫ x 2 e 3 x dx =
c. ∫ x e −2 x dx =
d. ∫ ln x dx =
e. ∫ x ln x dx =
f. ∫ x 3 ln x dx =
Hamilton Education Guides
2
Advanced Integration
1.1 Integration by Parts
h. ∫ e x sin x dx =
g. ∫ e 2 x cos 2 x dx =
i. ∫ x cos 3x dx =
Solutions:
a. Given ∫ xe x dx let u = x and dv = e x dx then du = dx and ∫ dv = ∫ e x dx which implies v = e x .
Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫ xe dx = xe − ∫ e dx = xe − e + c = e (x − 1) + c
x
x
x
x
x
x
Check: Let y = e x (x − 1) + c , then y ′ = e x ⋅ (x − 1) + 1 ⋅ e x + 0 = xe x − e x + e x = xe x − e x + e x = xe x
1
3
b. Given ∫ x 2 e 3 x dx let u = x 2 and dv = e 3 x dx then du = 2 x dx and ∫ dv = ∫ e 3 x dx which implies v = e 3 x .
Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫
x 2 e 3x 2
1 3x
1
x 2 e 3 x dx = x 2 ⋅ e 3 x −
−
x e 3 x dx
e ⋅ 2 x dx =
3
3
3
3
(1 )
∫
∫
To integrate ∫ x e 3 x dx let u = x and dv = e 3 x dx then du = dx and ∫ dv = ∫ e 3 x dx which implies
v=
1 3x
e . Therefore,
3
∫
xe 3 x e 3 x
1
1 3x
x e 3 x dx = x ⋅ e 3 x −
−
+c
e dx =
3
9
3
3
(2)
∫
Combining equations ( 1 ) and ( 2 ) together we obtain:
∫
x 2 e 3 x dx =
x 2 e 3x 2
−
3
3
∫
x e 3 x dx =

1 2 3x 2 3x 2 3x
x 2 e 3 x 2  xe 3 x e 3 x
x e − xe +
e +c
−
−
+ c =


9
27
3
3
3 3
9

(
) (
1
2
2
2 3x
2 x ⋅ e 3 x + 3e 3 x ⋅ x 2 − 1 ⋅ e 3 x + 3e 3 x ⋅ x
e + c , then y ′ =
3
9
9
27
2
2
2
2
2
+
⋅ 3e 3 x + 0 = xe 3 x + x 2 e 3 x − e 3 x − xe 3 x + e 3 x = x 2 e 3 x
27
9
3
3
9
1
3
Check: Let y = x 2 e 3 x − xe 3 x +
)
c. Given ∫ x e −2 x dx let u = x and dv = e −2 x dx then du = dx and ∫ dv = ∫ e −2 x dx which implies
v=−
∫xe
1 −2 x
. Using the integration by parts formula
e
2
−2 x
dx = x ⋅ −
1
1 −2 x 1
1
1
−2 x 1 −2 x
− e
+c
e
+
e − 2 x dx = − x e − 2 x +
e − 2 x dx = − x e
2
4
2
2
2
2
1
2
∫
∫
1
4
1
1
1
1 ⋅ e − 2 x − 2e − 2 x ⋅ x − ⋅ −2e − 2 x + 0 = − e − 2 x
2
4
2
Check: Let y = − x e −2 x − e −2 x + c , then y ′ = −
xe − 2 x +
∫ u dv = u v − ∫ v du we obtain
(
)
1 −2 x
= xe −2 x
e
2
Hamilton Education Guides
3
Advanced Integration
1.1 Integration by Parts
1
x
d. Given ∫ ln x dx let u = ln x and dv = dx then du = dx and ∫ dv = ∫ dx which implies v = x .
Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
1
∫ ln x dx = ln x ⋅ x − ∫ x ⋅ x dx = x ln x − ∫ dx = x ln x − x + c
1
x
Check: Let y = x ln x − x + c , then y ′ =  1 ⋅ ln x + ⋅ x  − 1 + 0 = ln x + 1 − 1 = ln x


1
2
1
x
e. Given ∫ x ln x dx let u = ln x and dv = x dx then du = dx and ∫ dv = ∫ x dx which implies v = x 2 .
Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
1
1
1
1
1
1
1
∫ x ln x dx = ln x ⋅ 2 x 2 − ∫ 2 x 2 ⋅ x dx = 2 x ln x − 2 ∫ x dx = 2 x ln x − 4 x + c
1
2
2
1
4
Check: Let y = x 2 ln x − x 2 + c , then y ′ =
2
2
1
1
1
1 2 1
 2 x ⋅ ln x + ⋅ x  − ⋅ 2 x + 0 = x ln x + x − x = x ln x
2
2
x
2
4

1
4
1
x
f. Given ∫ x 3 ln x dx let u = ln x and dv = x 3 dx then du = dx and ∫ dv = ∫ x 3 dx which implies v = x 4 .
Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
1
1
1
1
1
1 4
1
∫ x ln x dx = ln x ⋅ 4 x 4 − ∫ 4 x 4 ⋅ x dx = 4 x ln x − 4 ∫ x dx = 4 x ln x − 16 x + c
1
4
Check: Let y = x 4 ln x −
−
4
3
4
1
1
1
1 4
 1
x + c , then y ′ =  4 x 3 ⋅ ln x + ⋅ x 4  − ⋅ 4 x 3 + 0 = x 3 ln x + x 3
4
4
x
16
 16
1 3
x = x 3 ln x
4
g. Given ∫ e 2 x cos 2 x dx let u = e 2 x and dv = cos 2 x dx then du = 2e 2 x dx and ∫ dv = ∫ cos 2 x dx which
1
2
implies v = sin 2 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we
1
2
obtain ∫ e 2 x cos 2 x dx = e 2 x ⋅ sin 2 x −
e 2 x sin 2 x
1
sin 2 x ⋅ 2e 2 x dx =
− e 2 x sin 2 x dx
2
2
∫
∫
(1 )
To integrate ∫ e 2 x sin 2 x dx let u = e 2 x and dv = sin 2 x dx then du = 2e 2 x dx and ∫ dv = ∫ sin 2 x dx which
1
2
1
2
implies v = − cos 2 x . Thus, ∫ e 2 x sin 2 x dx = e 2 x ⋅ − cos 2 x +
∫
+ e 2 x cos 2 x dx
e 2 x cos 2 x
1
cos 2 x ⋅ 2e 2 x dx = −
2
2
∫
( 2 ) . Combining equations ( 1 ) and ( 2 ) together we obtain:
Hamilton Education Guides
4
Advanced Integration
1.1 Integration by Parts

e 2 x sin 2 x
e 2 x sin 2 x
e 2 x sin 2 x  e 2 x cos 2 x
− e 2 x sin 2 x dx =
− −
+ e 2 x cos 2 x dx  =


2
2
2
2


∫
e 2 x cos 2 x dx =
+
e 2 x cos 2 x
− e 2 x cos 2 x dx . Taking the
2
∫
∫
∫
∫ e cos 2 x dx from the right hand side of the equation
2x
to the left hand side we obtain ∫ e 2 x cos 2 x dx + ∫ e 2 x cos 2 x dx =
e 2 x sin 2 x e 2 x cos 2 x
+
and
2
2
∫
2 e 2 x cos 2 x dx =
Check: Let y =
−
(
e 2 x sin 2 x e 2 x cos 2 x
+
which implies
2
2
∫ e cos 2 x dx = 4 ( e
1
2x
2x
)
sin 2 x + e 2 x cos 2 x + c
)
2
2
2
1 2x
e sin 2 x + e 2 x cos 2 x + c , then y ′ = e 2 x ⋅ sin 2 x + cos 2 x ⋅ e 2 x + e 2 x ⋅ cos 2 x
4
4
4
4
2
4
2
2
4
sin 2 x ⋅ e 2 x + 0 = e 2 x sin 2 x + e 2 x cos 2 x − e 2 x sin 2 x = e 2 x cos 2 x = e 2 x cos 2 x
4
4
4
4
4
h. Given ∫ e x sin x dx let u = e x and dv = sin x dx then du = e x dx and ∫ dv = ∫ sin x dx which implies
v = − cos x . Using the integration by parts formula
∫ u dv = u v − ∫ v du we obtain
(1 )
∫ e sin x dx = e ⋅ − cos x − ∫ − cos x ⋅ e dx = − e cos x + ∫ e cos x dx
x
x
x
x
x
To integrate ∫ e x cos x dx let u = e x and dv = cos x dx then du = e x dx and ∫ dv = ∫ cos x dx which
(2)
implies v = sin x . Thus, ∫ e x cos x dx = e x ⋅ sin x − ∫ sin x ⋅ e x dx = e x sin x − ∫ e x sin x dx
Combining equations ( 1 ) and ( 2 ) together we obtain:
∫ e sin x dx = − e cos x + ∫ e cos x dx = − e cos x + e sin x − ∫ e sin x dx
x
x
x
x
x
x
Taking the ∫ e x sin x dx from the right hand side of the equation to the left hand side we obtain
∫ e sin x dx + ∫ e sin x dx = − e cos x + e sin x which implies 2∫ e sin x dx = − e cos x + e sin x
x
x
x
1
2
x
x
x
x
1
2
and ∫ e x sin x dx = − e x cos x + e x sin x + c
1
2
1
2
1
2
1
2
1
2
1
2
Check: Let y = − e x cos x + e x sin x + c , then y ′ = − e x ⋅ cos x + sin x ⋅ e x + e x ⋅ sin x + cos x ⋅ e x
1
2
1
2
1
2
1
2
= − e x cos x + e x sin x + e x sin x + e x cos x =
1 x
1
e sin x + e x sin x = e x sin x
2
2
i. Given ∫ x cos 3x dx let u = x and dv = cos 3x dx then du = dx and ∫ dv = ∫ cos 3x dx which implies
Hamilton Education Guides
5
Advanced Integration
v=
1.1 Integration by Parts
1
sin 3 x . Using the integration by parts formula
3
1
1
1
1
1
3
1
9
∫ u dv = u v − ∫ v du we obtain
∫ x cos 3x dx = x ⋅ 3 sin 3x − 3 ∫ sin 3x dx = 3 x sin 3 x + 9 cos 3 x + c
Check: Let y = x sin 3x + cos 3x + c , then y ′ =
+
1
( 1⋅ sin 3x + cos 3x ⋅ 3 ⋅ x ) − 1 ⋅ sin 3x ⋅ 3 + 0 = 1 sin 3x
3
9
3
1
3
x cos 3 x − sin 3 x = x cos 3 x
3
3
Example 1.1-2: Evaluate the following indefinite integrals:
x
3
a. ∫ x sec 2 5 x dx =
b. ∫ x sec 2 (x +1) dx =
d. ∫ arc sin 6 x dx =
e. ∫ arc sin y dy =
f. ∫ arc cos x dx =
h. ∫ arc tan 10 x dx =
i. ∫
c. ∫ sin 2 x dx =
1
5
x
2
g. ∫ arc cos dx =
x ex
(1 + x )2
dx =
Solutions:
a. Given ∫ x sec 2 5 x dx let u = x and dv = sec 2 5 x dx then du = dx and ∫ dv = ∫ sec 2 5 x dx which implies
v=
1
tan 5 x . Using the integration by parts formula
5
1
1
1
∫ u dv = u v − ∫ v du we obtain
1
∫ x sec 5x dx = x ⋅ 5 tan 5x − 5 ∫ tan 5x dx = 5 x tan 5 x − 25 ln sec 5 x + c
2
1
5
Check: Let y = x tan 5 x −
=
(
)
sec 5 x tan 5 x
1
1
⋅5 + 0
1 ⋅ tan 5 x + sec 2 5 x ⋅ 5 ⋅ x −
ln sec 5 x + c , then y ′ =
5
25 sec 5 x
25
tan 5 x 5 x sec 2 5 x 5 sec 5 x tan 5 x
tan 5 x
tan 5 x
=
= x sec 2 5 x
+
−
+ x sec 2 5 x −
5
5
25 sec 5 x
5
5
b. Given ∫ x sec 2 (x +1) dx let u = x and dv = sec 2 (x + 1) dx then du = dx and ∫ dv = ∫ sec 2 (x + 1) dx
which implies v = tan (x + 1) . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫ x sec (x +1) dx = x ⋅ tan (x + 1) − ∫ tan (x + 1) dx = x tan (x + 1) − ln sec (x + 1) + c
2
Check: Let y = x tan (x + 1) − ln sec (x + 1) + c , then y ′ = 1 ⋅ tan (x + 1) + sec 2 (x + 1) ⋅ x −
sec (x + 1) tan (x + 1)
+0
sec (x + 1)
= tan (x + 1) + x sec 2 (x + 1) − tan (x + 1) = x sec 2 (x + 1)
x
3
c. Given ∫ sin 2 x dx let u =
Hamilton Education Guides
x
dx
and dv = sin 2 x dx then du =
and
3
3
∫ dv = ∫ sin 2 x dx which implies
6
Advanced Integration
v=−
1.1 Integration by Parts
1
cos 2 x . Using the integration by parts formula
2
1
x
x
1
∫ u dv = u v − ∫ v du we obtain
1
dx
1
∫ 3 sin 2 x dx = 3 ⋅ − 2 cos 2 x + 2 ∫ cos 2 x 3 = − 6 x cos 2 x + 12 sin 2 x + c
1
1
1
sin 2 x + c , then y ′ = − ( 1 ⋅ cos 2 x − sin 2 x ⋅ 2 ⋅ x ) + cos 2 x ⋅ 2 + 0
12
12
6
1
1
1
x
= − cos 2 x + x sin 2 x + cos 2 x = sin 2 x
6
3
6
3
1
6
Check: Let y = − x cos 2 x +
6 dx
d. Given ∫ arc sin 6 x dx let u = arc sin 6 x and dv = dx then du =
1 − (6 x )2
and ∫ dv = ∫ dx which
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
6 dx
x dx
(1 )
∫ arc sin 6 x dx = arc sin 6 x ⋅ x − ∫ x ⋅ 1 − 36 x 2 = x arc sin 6 x − 6∫ 1 − 36 x 2
To integrate ∫
dx = −
= −
dw
72 x
x dx
1 − 36 x
2
Therefore, ∫
use the substitution method by letting w = 1 − 36 x 2 then
x dx
1 − 36 x 2
dw
1
= −
−
72
x
72
w
x
= ∫
(
1
dw
−1
dw
= −72 x and
dx
1
1− 1
1
∫ w = − 72 ∫ w dw = − 72 ⋅ 1 − 1 w
2
2
2
)
1
2 −1
1 12
1 2 1
1
1 1
w = −
1 − 36 x 2 2
⋅
w 2 = − ⋅ w2 = −
36
72 1
72 2−1
36
(2)
2
Combining equations ( 1 ) and ( 2 ) together we obtain
∫
arc sin 6 x dx = x arc sin 6 x − 6
Check: Let y = x arc sin 6 x +
= arc sin 6 x +
x dx
∫ 1 − 36 x 2
(
= x arc sin 6 x +
(
)
1
6x
1
1
−
1 − 36 x 2 2 + c , then y ′ = arc sin 6 x +
6
1 − 36 x 2 12
6x
1 − 36 x 2
−
6x
1 − 36 x 2
1
5
v = y . Using the integration by parts formula
1
1
dy
dy
1− y 2
72 x
1 − 36 x 2
+0
and ∫ dv = ∫ dy which implies
∫ u dv = u v − ∫ v du we obtain
1
1
y dy
∫ 5 arc sin y dy = 5 arc sin y ⋅ y − 5 ∫ y ⋅ 1 − y 2 = 5 y arc sin y − 5 ∫ 1 − y 2
Hamilton Education Guides
)
= arc sin 6 x
e. Given ∫ arc sin y dy let u = arc sin y and dv = dy then du =
1
(
)
1
1
1
6
1 − 36 x 2 2 + c = x arc sin 6 x + 1 − 36 x 2 2 + c
6
36
(1 )
7
Advanced Integration
To integrate ∫
dy = −
1
2
dw
2y
1.1 Integration by Parts
y dy
1− y 2
Therefore, ∫
2 −1
1
dw
= −2 y and
dy
use the substitution method by letting w = 1 − y 2 then
1 2
2 1
y dy
y
dw
1
= −
y
2
−
2
w
= ∫
1− y 2
(
1
1
)
dw
−1
1
1
1− 1
1
∫ w = − 2 ∫ w dw = − 2 ⋅ 1 − 1 w
2
2
2
1
(2)
= − ⋅ 2−1 w 2 = − ⋅ w 2 = − w 2 = − 1 − y 2 2
2
Combining equations ( 1 ) and ( 2 ) together we obtain:
∫
1
1
1
y arc sin y −
arc sin y dy =
5
5
5
1
5
Check: Let w = y arc sin y +
=
1
1
arc sin y +
5
5
y dy
∫ 1− y 2
(
=
(
)
1
1
1
y arc sin y + 1 − y 2 2 + c
5
5
)
1
1
1
1
1 − y 2 2 + c , then w′ = arc sin y +
5
5
5
y
1− y 2
−
1
5
y
1− y 2
y
1 − y2
1 1
⋅ ⋅
5 2
−
2y
1 − y2
1
arc sin y
5
=
f. Given ∫ arc cos x dx let u = arc cos x and dv = dx then du =
− dx
1− x 2
and ∫ dv = ∫ dx which implies
∫ u dv = u v − ∫ v du we obtain
v = x . Using the integration by parts formula
x dx
−dx
(1 )
∫ arc cos x dx = arc cos x ⋅ x − ∫ x ⋅ 1 − x 2 = x arc cos x + ∫ 1 − x 2
To integrate ∫
dx = −
1
2
dw
2x
x dx
1− x 2
use the substitution method by letting w = 1 − x 2 then
Therefore, ∫
2 −1
1
1 2
2 1
x dx
1− x 2
= ∫
dw
1
= −
2
w − 2x
x
(
1
1
)
dw
∫ w
= −
dw
= −2 x and
dx
−1
1− 1
1
1 1
w 2 dw = − ⋅
w 2
2
2 1− 1
∫
2
1
(2)
= − ⋅ 2−1 w 2 = − ⋅ w 2 = − w 2 = − 1 − x 2 2
2
+0
Combining equations ( 1 ) and ( 2 ) together we obtain:
∫
arc cos x dx = x arc cos x +
x dx
∫ 1− x 2
(
)
(
)
1
= x arc cos x − 1 − x 2 2 + c
1
Check: Let y = x arc cos x − 1 − x 2 2 + c , then y ′ = arc cos x −
−
x
1− x
2
+
Hamilton Education Guides
x
1− x 2
x
1− x
2
−
1
2
−2 x
1− x
2
+ 0 = arc cos x
= arc cos x
8
Advanced Integration
1.1 Integration by Parts
x
2
x
and dv = dx then du =
2
− dx
and ∫ dv = ∫ dx which
( )2
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
g. Given ∫ arc cos dx let u = arc cos
x
x
∫ arc cos 2 dx = arc cos 2 ⋅ x − ∫ x ⋅
To integrate ∫
x dx
−dx
()
2
1 − 2x
2
x
2
= x arc cos +
( )2
x dx
2
x
= ∫
(1 )
()
2
1 − 2x
()
2
1 − 2x
2
1
1
1
dw
2x 1
x
=−
⋅ =−
dx
2 2
2
−1
x −2dw
dw
= −2
x
w
w
∫
= − 2 ⋅ 2−1 w 2 = − 2 ⋅ w 2 = − 4w 2 = − 41 − (2x ) 
2
x dx
2
1 − 2x
2 −1
∫
use the substitution method by letting w = 1 − (2x ) then
and dx = − dw Therefore, ∫
1
1
2
2 1 − 2x
= − 2∫ w 2 dw = − 2 ⋅
1
1 − 12
1
2 2

1− 1
w
2
(2)

Combining equations ( 1 ) and ( 2 ) together we obtain:
∫
arc cos
x
x 1
dx = x arc cos +
2
2 2
x dx
∫
( )2
1 − 2x
1
()
1
()
2 2
x
x
Check: Let y = x arc cos − 2 1 − 2x  + c , then y ′ = arc cos −
2
2


x
2
= arc cos −
x
( )2
+
2 1 − 2x
x
1
()
2 2
2 2
x
x 4

= x arc cos − 1 − 2x  + c = x arc cos − 2 1 − 2x  + c
2
2 2



( )2
2 1 − 2x
= arc cos
x
( )2
−
2 1 − 2x
x
( )2
⋅−
1 − 2x
1
+0
2
x
2
h. Given ∫ arc tan 10 x dx let u = arc tan 10 x and dv = dx then du =
10 dx
1 + (10 x )2
and ∫ dv = ∫ dx which
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
10 dx
x dx
(1 )
∫ arc tan 10 x dx = arc tan 10 x ⋅ x − ∫ x ⋅ 1 + (10 x )2 = x arc tan 10 x − 10∫ 1 + (10 x )2
To integrate ∫
And dx =
dw
200 x
x dx
1 + (10 x )
2
Thus, ∫
use the substitution method by letting w = 1 + (10 x )2 then
x dx
1 + (10 x )
2
= ∫
x dw
1
=
w 200 x
200
dw
1
1
dw
= 200 x
dx
∫ w = 200 ln w = 200 ln 1 + (10 x )
2
(2)
Combining equations ( 1 ) and ( 2 ) together we obtain:
Hamilton Education Guides
9
Advanced Integration
1.1 Integration by Parts
x dx
1
∫ arc tan 10 x dx = x arc tan 10 x − 10∫ 1 + (10 x )2 = x arc tan 10 x − 20 ln 1 + (10 x ) + c
Check: Let y = x arc tan 10 x −
arc tan 10 x +
i. Given ∫
x ex
(1 + x )2
∫ (1 + x )2
=
1 + ( 10 x )
2
x
1
20 x
1
−
+0
ln 1 + (10 x )2 + c , then y ′ = arc tan 10 x +
2
20 1 + ( 10 x )2
20
1 + ( 10 x )
−
x
1 + ( 10 x )2
= arc tan 10 x
dx let u = xe x and dv = ( 1 + x )−2 dx then du = e x ( 1 + x ) dx and
which implies v = −
x ex
x
2
dx = xe x ⋅
1
. Using the integration by parts formula
1+ x
−1
−
1+ x
∫
∫ dv = ∫ ( 1 + x ) dx
−2
∫ u dv = u v − ∫ v du we obtain
− xe x + e x ( 1 + x )
−1
− xe x
− xe x
⋅ e x ( 1 + x ) dx =
+ e x dx =
+ex +c =
+c
( 1 + x)
1+ x
1+ x
1+ x
∫
ex
− xe x + e x + xe x
+c
+c =
1+ x
1+ x
Check: Let y =
e ( 1 + x) − e
ex
e x + xe x − e x
+0 =
+ c , then y ′ =
1+ x
( 1 + x )2
( 1 + x )2
x
x
=
xe x
( 1 + x )2
Example 1.1-3: Evaluate the following indefinite integrals:
a. ∫ sin 3 x dx =
b. ∫ sin 2 x dx =
c. ∫ arctan x dx =
d. ∫ sin ( ln x ) dx =
e. ∫ x 2 e x dx =
f. ∫ x 3 sin x dx =
g. ∫ x 2 cos 3x dx =
h. ∫ e − x cos x dx =
i. ∫ e −3 x sin 3x dx =
Solutions:
a. Given ∫ sin 3 x dx = ∫ sin 2 x ⋅ sin x dx let u = sin 2 x and dv = sin x dx then du = 2 sin x cos x dx and
∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du
we obtain ∫ sin 3 x dx = sin 2 x ⋅ − cos x + ∫ cos x ⋅ 2 sin x cos x dx = − sin 2 x cos x + 2∫ cos 2 x sin x dx
(1 )
To integrate ∫ cos 2 x sin x dx use the integration by parts method again, i.e., let u = cos 2 x and
dv = sin x dx then du = −2 sin x cos x dx and
∫ dv = ∫ sin x dx which implies v = − cos x . Therefore,
∫ cos x sin x dx = cos x ⋅ − cos x − ∫ cos x ⋅ 2 sin x cos x dx = − cos x − 2∫ cos x sin x dx . Taking the
2
2
3
2
integral − 2∫ cos 2 x sin x dx from the right hand side of the equation to the left side we obtain
Hamilton Education Guides
10
Advanced Integration
1.1 Integration by Parts
1
(2)
∫ cos x sin x dx + 2∫ cos x sin x dx = − cos x . Therefore, ∫ cos x sin x dx = − 3 cos 3 x
2
2
2
3
Combining equations ( 1 ) and ( 2 ) together we have
1
2
∫ sin x dx = − sin x cos x + 2∫ cos x sin x dx = − sin 2 x cos x + 2 ⋅ − 3 cos 3 x + c = − sin x cos x − 3 cos x + c
2
3
(
)
− 1 − cos 2 x cos x −
2
2
3
1
2
2


3
cos 3 x + c = − cos x +  cos 3 x − cos 3 x  + c = cos x − cos x + c
3
3
3


Note that another method of solving the above problem (as was shown in Section 4.3) is in the
following way:
∫ sin x dx = ∫ sin x ⋅ sin x dx = ∫ (1 − cos x )⋅ sin x dx let u = cos x , then dx = − sin x and dx = − sin x .
2
3
du
du
2
(
)
(
)
Therefore, ∫ sin 3 x dx = ∫ sin 2 x ⋅ sin x dx = ∫ 1 − cos 2 x ⋅ sin x dx = ∫ 1 − u 2 ⋅ sin x ⋅ −
(
)
= − ∫ 1 − u 2 du = ∫ u 2 −1 du =
1
3
1
1 3
3
u − u + c = cos x − cos x + c
3
3
Check: Let y = cos 3 x − cos x + c , then y ′ =
(
du
sin x
)
1
⋅ 3 cos 2 x ⋅ − sin x + sin x + 0 = − cos 2 x sin x + sin x
3
= sin x 1 − cos 2 x = sin x sin 2 x = sin 3 x
b. Given ∫ sin 2 x dx = ∫ sin x ⋅ sin x dx let u = sin x and dv = sin x dx then du = cos x dx and ∫ dv = ∫ sin x dx
which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫ sin x dx = sin x ⋅ − cos x + ∫ cos x ⋅ cos x dx = − sin x cos x + ∫ cos x dx = − sin x cos x + ∫ (1 − sin x ) dx
2
2
2
= − sin x cos x + x − ∫ sin 2 x dx . Taking the integral ∫ sin 2 x dx from the right hand side of the
equation to the left side we have ∫ sin 2 x dx + ∫ sin 2 x dx = − sin x cos x + x . Therefore,
∫
2 sin 2 x dx = − sin x cos x + x and
1
1
x
x
∫ sin x dx = − 2 sin x cos x + 2 + c = − 4 sin 2 x + 2 + c
2
or, we can solve the given integral in the following way:
1
1
1
1
1 sin 2 x
x
x
∫ sin x dx = ∫ 2 ( 1 − cos 2 x ) dx = 2 ∫ dx − 2 ∫ cos 2 x dx = 2 − 2 ⋅ 2 + c = − 4 sin 2 x + 2 + c
2
1
4
1
4
x
2
1
2
1
2
Check: Let y = − sin 2 x + + c , then y ′ = − cos 2 x ⋅ 2 + + 0 = − cos 2 x +
1
2
x
2
1
2
1
2
1
2
1
1
= ( 1 − cos 2 x ) = sin 2 x or,
2
2
1
2
1
2
Let y = − sin x cos x + + c , then y ′ = − cos x cos x + sin x sin x + + 0 = − cos 2 x + sin 2 x +
Hamilton Education Guides
1
2
11
Advanced Integration
= −
1.1 Integration by Parts
(
)
1
1
1
1
1
1
1 1
1
1 − sin 2 x + sin 2 x + = − + sin 2 x + sin 2 x + = sin 2 x + sin 2 x = sin 2 x
2
2
2
2
2
2
2 2
2
c. Given ∫ arctan x dx let u = arc tan x and dv = dx then du =
dx
1+ x 2
∫ u dv = u v − ∫ v du we obtain
v = x . Substituting the integral with its equivalent value
dx
and ∫ dv = ∫ dx which implies
x dx
(1 )
∫ x arctan x dx = arc tan x ⋅ x − ∫ x ⋅ 1 + x 2 = x arc tan x − ∫ 1 + x 2
To integrate ∫
Therefore, ∫
x dx
use the substitution method by letting w = 1 + x 2 then
1+ x
2
x dx
= ∫
1+ x
2
x dw
1
=
w 2x
2
dw
1
1
∫ w = 2 ln w = 2 ln 1 + x
dw
dw
= 2 x and dx =
2x
dx
(2)
2
Combining equations ( 1 ) and ( 2 ) together we obtain:
1
x dx
∫ x arctan x dx = x arc tan x − ∫ 1 + x 2 = x arc tan x − 2 ln 1 + x + c
2
1
2
Check: Let y = x arc tan x − ln 1 + x 2 + c , then y ′ = arc tan x +
+
x
1+ x 2
−
x
1+ x 2
x
1+ x
2
−
1 2x
+ 0 = arc tan x
2 1+ x 2
= arc tan x
d. Given ∫ sin ( ln x ) dx let u = sin ( ln x ) and dv = dx then du =
cos ( ln x )
dx and
x
∫ dv = ∫ dx which
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫ sin ( ln x ) dx = sin ( ln x )⋅ x − ∫ x ⋅
cos ( ln x )
dx = x sin ( ln x ) − cos ( ln x ) dx
x
(1 )
∫
To integrate ∫ cos ( ln x ) dx use the integration by parts formula again, i.e., let u = cos ( ln x ) and
dv = dx then du =
− sin ( ln x )
dx and
x
= cos ( ln x ) ⋅ x + ∫ x ⋅
∫ dv = ∫ dx which implies v = x . Therefore, ∫ cos ( ln x ) dx
sin ( ln x )
dx = x cos ( ln x ) + sin ( ln x ) dx
x
∫
(2)
Combining equations ( 1 ) and ( 2 ) together we have
∫ sin ( ln x ) dx = x sin ( ln x ) − ∫ cos ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) − ∫ sin ( ln x ) dx
Taking the integral − ∫ sin ( ln x ) dx from the right hand side of the equation to the left hand side
we obtain ∫ sin ( ln x ) dx + ∫ sin ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) + c Therefore,
Hamilton Education Guides
12
Advanced Integration
1.1 Integration by Parts
2 sin ( ln x ) dx = x sin ( ln x ) − x cos ( ln x ) + c and thus
∫
Check: Let y =
=
x
x
∫ sin ( ln x ) dx = 2 sin ( ln x ) − 2 cos ( ln x ) + c
sin ( ln x ) x cos ( ln x ) cos ( ln x ) x sin ( ln x )
x
x
sin ( ln x ) − cos ( ln x ) + c , then y ′ =
+
−
+
+0
2
2
2
2x
2
2x
sin ( ln x ) cos ( ln x ) cos ( ln x ) sin ( ln x )
sin ( ln x ) sin ( ln x )
=
= sin ( ln x )
+
−
+
+
2
2
2
2
2
2
e. Given ∫ x 2 e x dx let u = x 2 and dv = e x dx then du = 2 x dx and ∫ dv = ∫ e x dx which implies v = e x .
Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
(1 )
∫ x e dx = x ⋅ e − ∫ e ⋅ 2 x dx = x e − 2∫ x e dx
2
x
2
x
2 x
x
x
To integrate ∫ x e x dx use the integration by parts formula again, i.e., let u = x and dv = e x dx
Then du = dx and ∫ dv = ∫ dx which implies v = e x . Therefore,
∫ x e dx = x ⋅ e − ∫ e ⋅ dx = xe − ∫ e dx = xe − e
x
x
x
x
x
x
(2)
x
Combining equations ( 1 ) and ( 2 ) together we have
∫ x e dx = x e − 2∫ x e dx = x e − 2 (xe − e )+ c = x e − 2 xe + 2e + c
2
x
2 x
x
2 x
x
2 x
x
x
x
Check: Let y = x 2 e x − 2 xe x + 2e x + c , then y ′ = 2 xe x + x 2 e x − 2e x − 2 xe x + 2e x + 0 = x 2 e x
f. Given ∫ x 3 sin x dx let u = x 3 and dv = sin x dx then du = 3x 2 dx and ∫ dv = ∫ sin x dx which implies
v = − cos x . Using the integration by parts formula
∫ u dv = u v − ∫ v du we obtain
(1 )
∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ 3x dx = − x cos x + 3∫ x cos x dx
3
3
3
2
2
To integrate ∫ x 2 cos x dx use the integration by parts formula again, i.e., let u = x 2 and dv = cos x dx
then du = 2 x dx and ∫ dv = ∫ cos x dx which implies v = sin x . Using the integration by parts
formula ∫ u dv = u v − ∫ v du we obtain
∫ x cos x dx = x ⋅ sin x − ∫ sin x ⋅ 2 x dx = x sin x − 2∫ x sin x dx
2
2
2
(2)
To integrate ∫ x sin x dx use the integration by parts formula again, i.e., let u = x and dv = sin x dx
then du = dx and ∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts
Hamilton Education Guides
13
Advanced Integration
1.1 Integration by Parts
formula ∫ u dv = u v − ∫ v du we obtain
(3 )
∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ dx = − x cos x + sin x
Combining equations ( 1 ) , ( 2 ) and ( 3 ) together we have
3
2
∫ x sin x dx = − x cos x + 3∫ x cos x dx = − x cos x + 3x sin x − 6∫ x sin x dx = − x cos x + 3 x sin x
3
3
2
3
2
+ 6 x cos x − 6 sin x + c
(
Check: Let y = − x 3 cos x + 3x 2 sin x + 6 x cos x − 6 sin x + c , then y ′ = − 3x 2 cos x + x 3 sin x
(
)
)
+ 6 x sin x + 3 x 2 cos x + (6 cos x − 6 x sin x ) − 6 cos x + 0 = x 3 sin x
g. Given ∫ x 2 cos 3x dx let u = x 2 and dv = cos 3x dx then du = 2 x dx and ∫ dv = ∫ cos 3x dx which
implies v =
sin 3 x
. Using the integration by parts formula
3
sin 3 x
sin 3 x
∫ u dv = u v − ∫ v du we obtain
2
1
(1 )
∫ x cos 3x dx = x 2 ⋅ 3 − ∫ 3 ⋅ 2 x dx = 3 x 2 sin 3x − 3 ∫ x sin 3x dx
2
To integrate ∫ x sin 3x dx use the integration by parts formula again, i.e., let u = x and dv = sin 3x dx
then du = dx and ∫ dv = ∫ sin 3x dx which implies v =
− cos 3 x
. Using the integration by parts
3
formula ∫ u dv = u v − ∫ v du we obtain
∫ x sin 3x dx = x ⋅
− cos 3 x
cos 3 x
1
1
⋅ dx = − x cos 3 x +
cos 3 x dx
+
3
3
3
3
∫
(2)
∫
Combining equations ( 1 ) and ( 2 ) together we have
1
2
1
2
1
2 1
∫ x cos 3x dx = 3 x 2 sin 3x − 3 ∫ x sin 3x dx = 3 x 2 sin 3x − 3 ⋅ − 3 x cos 3x − 3 ⋅ 3 ∫ cos 3x dx
2
=
1 2
2
2
x sin 3 x + x cos 3 x −
sin 3 x + c
3
9
27
1
3
2
9
(
)
2
1
sin 3 x + c , then y ′ =
2 x sin 3 x + 3 x 2 cos 3 x
27
3
6
2
2
2
2
2
3
+ ( cos 3 x − 3 x sin 3 x ) −
cos 3 x + c = x sin 3 x + x 2 cos 3 x + cos 3 x − x sin 3 x − cos 3 x
9
27
3
9
3
3
9
Check: Let y = x 2 sin 3x + x cos 3x −
=
3 2
x cos 3 x = x 2 cos 3 x
3
h. Given ∫ e − x cos x dx let u = cos x and dv = e − x dx then du = − sin x dx and ∫ dv = ∫ e − x dx which
Hamilton Education Guides
14
Advanced Integration
1.1 Integration by Parts
implies v = −e − x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫e
−x
∫
(1 )
∫
cos x dx = cos x ⋅ −e − x − e − x ⋅ sin x dx = − e − x cos x − e − x sin x dx
To integrate ∫ e − x sin x dx use the integration by parts formula again, i.e., let u = sin x and
dv = e − x dx then du = cos x dx and
∫e
−x
∫ dv = ∫ e dx which implies v = −e . Therefore,
−x
−x
∫
(2)
∫
sin x dx = sin x ⋅ −e − x + e − x ⋅ cos x dx = − e − x sin x + e − x cos x dx
Combining equations ( 1 ) and ( 2 ) together we have
∫e
−x
∫
∫
cos x dx = − e − x cos x − e − x sin x dx = − e − x cos x + e − x sin x − e − x cos x dx
Taking the integral
∫e
−x
cos x dx from the right hand side of the equation to the left hand side
we obtain ∫ e − x cos x dx + ∫ e − x cos x dx = − e − x cos x + e − x sin x therefore
∫
2 e − x cos x dx = − e − x cos x + sin x e − x and thus
∫e
−x
cos x dx = −
1 −x
1
e cos x + e − x sin x + c
2
2
(
) (
)
− e− x cos x e− x sin x
1
1
+
+ c , then y ′ = − − e − x cos x − e − x sin x + − e − x sin x + e − x cos x + 0
2
2
2
2
1 −x
1 −x
1 −x
1 −x
1 −x
1 −x
= e cos x + e sin x − e sin x + e cos x = e cos x + e cos x = e − x cos x
2
2
2
2
2
2
Check: Let y =
i. Given ∫ e −3 x sin 3x dx let u = sin 3x and dv = e −3 x dx then du = 3 cos 3x dx and ∫ dv = ∫ e −3 x dx which
1
3
implies v = − e −3 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫e
−3 x
1
1
1 −3 x
sin 3 x dx = sin 3 x ⋅ − e −3 x +
⋅ 3 cos 3 x dx = − e −3 x sin 3 x + e −3 x cos 3 x dx
e
3
3
3
∫
∫
(1 )
To integrate ∫ e −3 x cos 3x dx use the integration by parts formula again, i.e., let u = cos 3x and
dv = e −3 x dx then du = − 3 sin 3 x and
∫e
−3 x
∫ dv = ∫ e
−3 x
1
dx which implies v = − e −3 x . Therefore,
3
1
1
1 −3 x
cos 3 x dx = cos 3 x ⋅ − e −3 x −
e
⋅ 3 sin 3 x dx = − e −3 x cos 3 x − e −3 x sin 3 x dx
3
3
3
∫
∫
(2)
Combining equations ( 1 ) and ( 2 ) together we have
∫e
−3 x
1
1
1
sin 3 x dx = − e −3 x sin 3 x + e −3 x cos 3 x dx = − e −3 x sin 3 x − e −3 x cos 3 x − e −3 x sin 3 x dx
3
3
3
Taking the integral
∫
∫e
Hamilton Education Guides
−3 x
∫
sin 3 x dx from the right hand side of the equation to the left hand side
15
Advanced Integration
1.1 Integration by Parts
1
3
1
3
we obtain ∫ e −3 x sin 3x dx + ∫ e −3 x sin 3x dx = − e −3 x sin 3x − e −3 x cos 3x therefore
1
1
2 e −3 x sin 3 x dx = − e −3 x sin 3 x − e −3 x cos 3 x and thus
3
3
∫
1
6
1
6
∫e
−3 x
sin 3 x dx = −
1
1 −3 x
e
sin 3 x − e − 3 x cos 3 x
6
6
(
)
1
− 3e −3 x sin 3 x + 3e −3 x cos 3 x
6
1
1
1
1
1
− − 3e −3 x cos 3 x − 3e −3 x sin 3 x + 0 = e −3 x sin 3 x − e −3 x cos 3 x + e −3 x cos 3 x + e −3 x sin 3 x
2
2
2
2
6
1
1
= e −3 x sin 3x + e −3 x sin 3x = e −3 x sin 3x
2
2
Check: Let y = − e − 3 x sin 3x − e − 3 x cos 3x , then y ′ = −
(
)
Example 1.1-4: Evaluate the following indefinite integrals:
x
3
a. ∫ (x + 1)4 dx =
b. ∫ (x − 3)(3x − 1)3 dx =
c. ∫ (x + 1) csc 2 x dx =
d. ∫ x sec 2 x dx =
e. ∫ x x − 5 dx =
f. ∫ ln x 2 +1 dx =
h. ∫ cos −1 3x dx =
i. ∫ tan −1 5 x dx =
1
5
g. ∫ e x sin 3x dx =
(
)
Solutions:
x
3
a. Given ∫ (x + 1)4 dx =
1
x (x + 1)4 dx let u = x and dv = (x + 1)4 dx then du = dx and
3
∫
∫ dv = ∫ (x + 1) dx
4
1
5
which implies v = (x + 1)5 . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
x (x + 1)5
1
1
−
x (x + 1)4 dx = ⋅
3
5
3⋅5
3
∫
Check: Let y =
+
x (x + 1)
1 1
5
− ⋅ (x + 1)5+1 + c =
∫ (x + 1) ⋅ dx =
5
15
15 6
[
x ( x + 1) 5 1
( x + 1) 6 + c
−
15
90
]
5
x (x + 1)5 1
(x + 1)6 + c , then y ′ = 1 (x + 1)5 + 5 x(x + 1)4 − 6 (x + 1)5 + 0 = (x + 1)
−
15
90
15
15
90
5
5x
(x + 1)4 − (x + 1) = 5 x (x + 1)4 = x (x + 1)4
15
15
3
15
b. Given ∫ (x − 3)(3x − 1)3 dx let u = x − 3 and dv = (3x − 1)3 dx then du = dx and ∫ dv = ∫ (3x − 1)3 dx
which implies v =
1
(3x − 1)4 . Using the integration by parts formula
12
∫ u dv = u v − ∫ v du we obtain
(3x − 1) − 1 (3x − 1)4 dx (x − 3) (3x − 1) − 1 ⋅ 1 (3x − 1)4+1 + c
3
=
∫ (x − 3)(3x − 1) dx = (x − 3)⋅
∫
4
12
=
4
12
12
12 15
( x − 3)(3 x − 1) 4 − 1 (3 x − 1) 5 + c
12
180
Hamilton Education Guides
16
Advanced Integration
Check: Let y =
=
1.1 Integration by Parts
(x − 3)(3x − 1)4 − 1 (3x − 1)5 + c , then y ′ = 1 [(3x − 1)4 + 12(x − 3)(x + 1)3 ]− 15 (3x − 1)4 + 0
180
12
12
180
3
1
(3x − 1)4 + 12(x − 3)(x + 1) − 1 (3x − 1)4 = (x − 3)(x + 1)3
12
12
12
c. Given ∫ (x + 1) csc 2 x dx let u = x + 1 and dv = csc 2 x dx then du = dx and ∫ dv = ∫ csc 2 x dx which
implies v = − cot x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
∫ (x + 1) csc x dx = (x + 1)⋅ − cot x + ∫ cot x dx = − (x + 1)cot x + ln sin x + c
2
[
x
] cos
+ 0 = − cot x
sin x
Check: Let y = −(x + 1) cot x + ln sin x + c , then y ′ = − cot x − (x + 1) csc 2 x +
+ (x + 1) csc 2 x + cot x = (x + 1) csc 2 x
d. Given ∫ x sec 2 x dx let u = x and dv = sec 2 x dx then du = dx and ∫ dv = ∫ sec 2 x dx which implies
v = tan x . Using the integration by parts formula
∫ u dv = u v − ∫ v du we obtain
∫ x sec x dx = x ⋅ tan x − ∫ tan x dx = x tan x − ln sec x + c
2
Check: Let y = x tan x − ln sec x + c , then y ′ = tan x + x sec 2 x −
sec x tan x
+ 0 = tan x + x sec 2 x − tan x
sec x
= x sec 2 x
e. Given ∫ x x − 5 dx let u = x and dv = x − 5 dx then du = dx and ∫ dv = ∫ x − 5 dx which implies
v=
3
2
(x − 5) 2 . Using the integration by parts formula
3
2
3
2
2
3
2
2
3
2
∫ u dv = u v − ∫ v du we obtain
2
1
∫ x x − 5 dx = x ⋅ 3 (x − 5) − ∫ 3 (x − 5) dx = 3 x (x − 5) − 3 ⋅ 1 + 3 (x − 5)
3 +1
2
+c
2
−
=
3
2
x ( x − 5) 2
3
3
5
5
4
2
2 2
x ( x − 5) 2 − ( x − 5) 2 + c
⋅ ( x − 5) 2 + c =
3
15
3 5
3
3
3
5
1
2
2 3
4 5
2
4
x (x − 5) 2 − (x − 5) 2 + c , then y ′ = (x − 5) 2 + ⋅ x(x − 5) 2 − ⋅ (x − 5) 2 + 0
3
3 2
15 2
15
3
1
3
3
1
2
2
= ( x − 5) 2 + x ( x − 5) 2 − ( x − 5) 2 = x ( x − 5) 2 = x x − 5
3
3
Check: Let y =
(
)
(
)
f. Given ∫ ln x 2 +1 dx let u = ln x 2 + 1 and dv = dx then du =
v = x . Using the integration by parts formula
Hamilton Education Guides
2x
2
x +1
dx and
∫ dv = ∫ dx which implies
∫ u dv = u v − ∫ v du we obtain
17
Advanced Integration
∫ (
)
1.1 Integration by Parts
) ∫ x 2 x+ 1 dx = x ln ( x 2 + 1)− 2∫ x 2x + 1 dx = x ln ( x + 1)− 2∫ 1 − 1  dx
 x +1
(
ln x 2 +1 dx = ln x 2 + 1 ⋅ x − x ⋅
(
)
= x ln x 2 + 1 − 2∫ dx + 2∫
1
2
x +1
(
2
2
2
(
)
−1
2
dx = x ln x + 1 − 2 x + 2 tan x + c
(
)
)
Check: Let y = x ln x 2 + 1 − 2 x + 2 tan −1 x + c , then y ′ = 1 ⋅ ln x 2 + 1 +
(
2x
2
x +1
⋅x−2+
2
1+ x 2
) x2 x+ − 2 + x 2+ = ln (x + 1)+ 2 x − 2 x − 2 + 2 = ln (x + 1)+ 0
= ln x 2 + 1 +
g. Given
2
2
2
2
1
2
2
2
2
x 2 +1
1
1
e x sin 3 x dx let u = e x and dv = sin 3 x dx then du = e x dx and
5
+0
x 2 +1
(
)
= ln x 2 + 1
∫ dv = ∫ sin 3x dx which
∫
1
3
implies v = − cos 3x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
1
1
1
1
1
1
e x sin 3 x dx = e x ⋅ − cos 3 x +
cos 3 x ⋅ e x dx = − e x cos 3 x +
e x cos 3 x dx
5
5
3
15
15
15
∫
∫
(1 )
∫
To integrate ∫ e x cos 3x dx use the integration by parts method again, i.e., let u = e x and
dv = cos 3 x dx then du = e x dx and
1
∫ dv = ∫ cos 3x dx which implies v = 3 sin 3x . Therefore,
1
1
1
1
(2)
∫ e cos 3x dx = e x ⋅ 3 sin 3x − 3 ∫ sin 3x ⋅ e x dx = 3 e x sin 3x − 3 ∫ e x sin 3x dx
x
Combining equations ( 1 ) and ( 2 ) together we have
1
1
1
1
1 x
1
e x sin 3 x dx = − e x cos 3 x +
e x cos 3 x dx = − e x cos 3 x +
e x sin 3 x dx
e sin 3 x −
5
15
15
45
45
15
∫
∫
Taking the integral −
obtain
= −
∫
1
e x sin 3 x dx from the right hand side of the equation to the left side we
45
∫
2
1
1
1
1 x
e x sin 3 x dx
e x sin 3 x dx +
e x sin 3 x dx = − e x cos 3 x +
e sin 3 x + c . Therefore,
9
5
15
45
45
∫
∫
∫
1 x
1 x
e cos 3 x +
e sin 3 x + c which implies
15
45
3
1
∫ e sin 3x dx = − 10 e cos 3 x + 10 e sin 3 x + c
x
x
x
3 x
1
3
3
1
e cos 3 x + e x sin 3 x + c , then y ′ = − e x ⋅ cos 3 x + sin 3 x ⋅ 3 ⋅ e x + e x ⋅ sin 3 x
10
10
10
10
10
9
3
9
1
3
1
e x sin 3 x
+ cos 3 x ⋅ 3 ⋅ e x + 0 = − e x cos 3 x + e x sin 3 x + e x sin 3 x + e x cos 3 x =
10
10
10
10
10
10
1
10 x
9 +1 x
+ e x sin 3 x =
e sin 3 x = e x sin 3 x
e sin 3 x =
10
10
10
Check: Let y = −
h. Given ∫ cos −1 3x dx let u = cos −1 3x and dv = dx then du = −
Hamilton Education Guides
3
1 − 9x 2
dx and
∫ dv = ∫ dx which
18
Advanced Integration
1.1 Integration by Parts
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
3 dx
3x
∫ cos 3x dx = cos 3x ⋅ x + ∫ x ⋅ 1 − 9 x 2 = x cos 3x + ∫ 1 − 9 x 2 dx
−1
−1
3x
To integrate ∫
implies dx = −
1
6
1 − 9x
dx use the substitution method by letting w = 1 − 9 x 2 then
2
dw
. Thus,
18 x
1
= − ⋅ 2w 2 = −
−1
1 − 9x 2
3
3x
∫ 1 − 9x 2
dx =
3x
∫ w
⋅−
1
dw
= −
6
18 x
and ∫ cos −1 3x dx = x cos −1 3x + ∫
Check: Let y = x cos −1 3x −
1 − 9x 2
3
1
6
∫
1
∫ w
= −
3x
dx = x cos−1 3 x −
1 − 9x 2
3x
+ c , then y ′ = cos −1 3 x −
i. Given ∫ tan −1 5 x dx let u = tan −1 5 x and dv = dx then du =
dw
2
1 − 9x
5
+
dw = −
1
w2
18 x
6 1 − 9x
dx and
1 + 25 x 2
dw
= −18 x which
dx
2
−1
1
w 2 dw
6
∫
1 − 9 x2
+c
3
+ 0 = cos −1 3 x
∫ dv = ∫ dx which
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
5 dx
5x
∫ tan 5x dx = tan 5x ⋅ x − ∫ x ⋅ 1 + 25x 2 = x tan 5x − ∫ 1 + 25x 2 dx
−1
−1
To integrate ∫
implies dx =
5x
1 + 25 x
2
−1
dx use the substitution method by letting w = 1 + 25 x 2 then
dw
. Thus,
50 x
5x
5 x dw
5x
−1
1
dw
dw
= 50 x which
dx
1
1
∫ 1 + 25x 2 dx = ∫ w ⋅ 50 x = 10 ∫ w = 10 ln w = 10 ln 1 + 25x and
1
2
∫ tan 5x dx = x tan 5x − ∫ 1 + 25x 2 dx = x tan 5 x − 10 ln 1 + 25 x + c
−1
−1
Check: Let y = x tan −1 5 x −
2
5x
1 50 x
1
−
+ 0 = tan −1 5 x
ln 1 + 25 x 2 + c , then y ′ = tan −1 5 x +
2 10
2
10
1 + 25 x
1 + 25 x
Example 1.1-5: Evaluate the following indefinite integrals:
a. ∫ sinh −1 5 x dx =
b. ∫ x tan −1 x dx =
d. ∫ cos 5 x cos 7 x dx =
e. ∫ e − x dx =
c. ∫ sin x sin 7 x dx =
x
5
f. ∫ x sinh 3x dx =
Solutions:
a. Given ∫ sinh −1 5 x dx let u = sinh −1 5 x and dv = dx then du =
5
1 + 25 x 2
dx and
∫ dv = ∫ x dx which
implies v = x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
Hamilton Education Guides
19
Advanced Integration
1.1 Integration by Parts
5 dx
5 x dx
(1 )
∫ sinh 5x dx = sinh 5x ⋅ x − ∫ x ⋅ 1 + 25x 2 = x sinh 5x − ∫ 1 + 25x 2
−1
−1
5 x dx
To get the integral of ∫
1 + 25 x
dw = 50 x dx which implies dx =
=
1
10
∫
1
1
w2
dw =
−1
use the substitution method by letting w = 1 + 25 x 2 then
2
dw
. Therefore,
50 x
5 x dx
5x
dw
5
1
∫ 1 + 25x 2 = ∫ w ⋅ 50 x = 50 ∫ w dw
−1
1− 1
1
2 12
1 1
1
1
1
w 2 dw =
w = w2 =
w 2 =
⋅
1 + 25 x 2
10
10
5
10 1 − 1
5
∫
(2)
2
Combining equations ( 1 ) and ( 2 ) together we have
∫
sinh −1 5 x dx = x sinh −1 5 x −
Check: Let y = x sinh −1 5 x −
sinh −1 5 x +
5x
1 + 25 x
2
5 x dx
∫ 1 + 25x 2
(
= x sinh −1 5 x −
(
)
1
1
1 + 25 x 2 2 + c
5
)
1
1
50 x
5x
1 + 25 x 2 2 + c , then y ′ = sinh −1 5 x +
+0
−
5
1 + 25 x 2 10 1 + 25 x 2
−
5x
1 + 25 x
2
= sinh −1 5 x
b. Given ∫ x tan −1 x dx let u = tan −1 x and dv = x dx then du =
1
1+ x 2
dx and
∫ dv = ∫ x dx which
1
2
implies v = x 2 . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
x2 1
−
2 2
dx
1 2
x2
1
1
1
x tan −1 x −
dx = x 2 tan −1 x −
2
2 1+ x 2
2
2
∫
x tan −1 x dx = tan −1 x ⋅
=
1
1
1 2
1
1 2
1
1
x tan −1 x − x + tan −1 x + c
dx =
x tan −1 x −
dx +
2
2
2
2
2 1+ x
2
2
∫
Check: Let y =
∫
x2 ⋅
1+ x
=
2
∫


1
∫ 1 − 1 + x 2 dx
∫
1
1 x2
1 1
1
1 2
1
1
− + ⋅
+0
x tan −1 x − x + tan −1 x + c , then y ′ = ⋅ 2 x ⋅ tan −1 x +
2
2 1+ x 2 2 2 1+ x 2
2
2
2
1 x2
1
1
1
1 x 2 +1 1
1 1
x tan −1 x + ⋅
+ ⋅
− = x tan −1 x + ⋅
− = x tan −1 x + − = x tan −1 x
2 1+ x 2 2 1+ x 2 2
2 1+ x 2 2
2 2
c. Given ∫ sin x sin 7 x dx let u = sin x and dv = sin 7 x dx then du = cos x dx and ∫ dv = ∫ sin 7 x dx which
1
7
implies v = − cos 7 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
1
1
1
1
∫ sin x sin 7 x dx = sin x ⋅ − 7 cos 7 x + 7 ∫ cos 7 x ⋅ cos x dx = − 7 sin x cos 7 x + 7 ∫ cos x ⋅ cos 7 x dx
(1 )
To integrate ∫ cos x ⋅ cos 7 x dx use the integration by parts method again, i.e., let u = cos x and
Hamilton Education Guides
20
Advanced Integration
1.1 Integration by Parts
dv = cos 7 x dx then du = − sin x dx and
1
1
∫ dv = ∫ cos 7 x dx which implies v = 7 sin 7 x . Therefore,
1
1
1
(2)
∫ cos x ⋅ cos 7 x dx = cos x ⋅ 7 sin 7 x + 7 ∫ sin 7 x ⋅ sin x dx = 7 cos x sin 7 x + 7 ∫ sin x sin 7 x dx
Combining equations ( 1 ) and ( 2 ) together we have
1
1
1
1
1
∫ sin x sin 7 x dx = − 7 sin x cos 7 x + 7 ∫ cos x ⋅ cos 7 x dx = − 7 sin x cos 7 x + 49 cos x sin 7 x + 49 ∫ sin x sin 7 x dx
Taking the integral
1
sin x sin 7 x dx to the left hand side and simplifying we have
49
∫
1 49
7
1 49
1
∫ sin x sin 7 x dx = − 7 ⋅ 48 sin x cos 7 x + 49 ⋅ 48 cos x sin 7 x + c = − 48 sin x cos 7 x + 48 cos x sin 7 x + c
Check: Let y = −
7
49
1
7
cos x ⋅ cos 7 x +
sin 7 x ⋅ sin x
cos x sin 7 x + c , then y ′ = −
sin x cos 7 x +
48
48
48
48
−
49
1
7
1
49 − 1
sin 7 x ⋅ sin x −
sin x ⋅ sin 7 x =
sin 7 x ⋅ sin x
cos 7 x ⋅ cos x + 0 =
sin x ⋅ sin 7 x +
48
48
48
48
48
=
48
sin 7 x ⋅ sin x = sin 7 x sin x
48
d. Given ∫ cos 5 x cos 7 x dx let u = cos 5 x and dv = cos 7 x dx then du = −5 sin 5 x dx and ∫ dv = ∫ cos 7 x dx
1
7
which implies v = sin 7 x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
5
1
5
1
∫ cos 5x cos 7 x dx = cos 5x ⋅ 7 sin 7 x + 7 ∫ sin 7 x ⋅ sin 5x dx = 7 cos 5x sin 7 x + 7 ∫ sin 5x ⋅ sin 7 x dx
(1 )
To integrate ∫ sin 5 x ⋅ sin 7 x dx use the integration by parts method again, i.e., let u = sin 5 x and
dv = sin 7 x dx then du = 5 cos 5 x dx and
1
1
∫ dv = ∫ sin 7 x dx which implies v = − 7 cos 7 x . Therefore,
1
5
5
∫ sin 5x ⋅ sin 7 x dx = sin 5x ⋅ − 7 cos 7 x + 7 ∫ cos 7 x ⋅ cos 5x dx = − 7 sin 5x cos 7 x + 7 ∫ cos 5x cos 7 x dx
(2)
Combining equations ( 1 ) and ( 2 ) together we have
1
5
1
5
25
∫ cos 5x cos 7 x dx = 7 cos 5x sin 7 x + 7 ∫ sin 5x ⋅ sin 7 x dx = 7 cos 5x sin 7 x − 49 sin 5x cos 7 x + 49 ∫ cos 5x cos 7 x dx
Taking the integral
25
cos 5 x cos 7 x dx to the left hand side and simplifying we have
49
∫
1 49
5 49
7
5
∫ cos 5x cos 7 x dx = 7 ⋅ 24 cos 5x sin 7 x − 49 ⋅ 24 sin 5x cos 7 x + c = 24 cos 5 x sin 7 x − 24 sin 5 x cos 7 x + c
Check: Let y =
7
5
35
49
cos 5 x sin 7 x −
sin 5 x cos 7 x + c , then y ′ = −
sin 5 x ⋅ sin 7 x +
cos 7 x ⋅ cos 5 x
24
24
24
24
Hamilton Education Guides
21
Advanced Integration
1.1 Integration by Parts
−
49 − 25
49
25
35
25
cos 5 x cos 7 x −
cos 5 x cos 7 x =
cos 5 x cos 7 x
sin 5 x ⋅ sin 7 x + 0 =
cos 5 x ⋅ cos 7 x +
24
24
24
24
24
=
24
cos 5 x cos 7 x = cos 5 x cos 7 x
24
x
5
e. Given ∫ e − x dx let u =
x
1
and dv = e − x dx then du = dx and
5
5
v = −e − x . Using the integration by parts formula
∫
−x
dx which implies
∫ u dv = u v − ∫ v du we obtain
−x
e
xe − x e − x
x
dx
x −x
−
+c = −
e dx = ⋅ −e − x + e − x ⋅
= −
5
5
5
5
5
5
∫
Check: Let y = −
∫ dv = ∫ e
( x + 1) + c
e − x xe − x e − x
xe − x e − x
xe − x
+
+
+0 =
−
+ c , then y ′ = −
5
5
5
5
5
5
f. Given ∫ x sinh 3x dx let u = x and dv = sinh 3x dx then du = dx and ∫ dv = ∫ sinh 3x dx which
1
3
implies v = cosh 3x dx . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
1
1
1
1
1
1
∫ x sinh 3x dx = x ⋅ 3 cosh 3x − ∫ 3 cosh 3x ⋅ dx = 3 x cosh 3x − 3 ∫ cosh 3x ⋅ dx = 3 x cosh 3 x − 9 sinh 3 x + c
1
9
1
3
Check: Let y = x cosh 3x − sinh 3x + c , then y ′ =
1
1
1
1
cosh 3 x + ⋅ 3 x sinh 3 x − ⋅ 3 cosh 3 x + 0 = cosh 3 x
3
9
3
3
1
+ x sinh 3 x − cosh 3 x = x sinh 3 x
3
Section 1.1 Practice Problems – Integration by Parts
1. Evaluate the following integrals using the integration by parts method.
x
2
a. ∫ xe 4 x dx =
b. ∫ cos x dx =
c. ∫ ( 5 − x ) e 5 x dx =
d. ∫ x sin 5 x dx =
e. ∫ x 3 − x dx =
f. ∫ x 3 e 3 x dx =
g. ∫ cos ( ln x ) dx =
h. ∫ tan −1 x dx =
i. ∫ ln x5dx =
j. ∫ x e − ax dx =
k. ∫ e x sin 3x dx =
l. ∫ e x cos 5 x dx =
x
3
2. Evaluate the following integrals using the integration by parts method.
a. ∫ x sec 2 x dx =
b. ∫ arc sin 3 y dy =
c. ∫ arc tan x dx =
d. ∫ sin 3 5 x dx =
e. ∫ x 2 cos x dx =
f. ∫ e −2 x cos 3x dx =
g. ∫ x ( 5 x − 1) 3 dx =
h. ∫ x csc 2 x dx =
i. ∫ cos −1 5 x dx =
j. ∫ sinh −1 x dx =
k. ∫ x sec 2 10 x dx =
l. ∫ sinh 7 x dx =
Hamilton Education Guides
2
3
x
5
22
Advanced Integration
1.2
1.2 Integration Using Trigonometric Substitution
Integration Using Trigonometric Substitution
Many integration problems involve radical expressions of the form
a2 − b2 x2
a2 + b2 x2
b2 x2 − a2
In such instances we can use a trigonometric substitution by letting
x=
a
sin t
b
x=
respectively to obtain
tan 2 t
= a 2 + a 2 tan 2 t = a
(1+ tan t ) = a sec t = a sec t
sec 2 t − a 2
= a 2 sec 2 t − a 2 = a
( sec t − 1) = a tan t = a tan t
a2 + b2 x2 =
a2 + b2 ⋅
b2 x2 − a2 =
b2 ⋅
b
a
sec t
b
(1− sin t ) = a cos t = a cos t
a2 − b2 ⋅
2
x=
a 2 − a 2 sin 2 t = a
a2 − b2 x2 =
a2
a
tan t
b
a2
b
2
a2
b2
sin 2 t =
2
2
2
2
2
2
Notice that using trigonometric substitution result in elimination of the radical expression. This
in effect reduces the difficulty of solving integrals with radical expressions.
Reminder 1:
bx
a
b
a
a
−1 b x
Given x = tan t , then t = tan
b
a
a
−1 b x
Given x = sec t , then t = sec
b
a
Given x = sin t , then t = sin −1
for −1 ≤ x ≤ 1 and −
for all x and −
π
2
π
2
t
≤t ≤
π
2
π
2
for x ≥ 1 or x ≤ −1 and 0 ≤ t
π
2
or π ≤ t
3π
2
Reminder 2:
In solving this class of integrals the integrand in the original variable may be obtained by the use
of a right triangle. For example, in a right triangle
•
sin t =
opposite
x
= . Therefore, using the Pythagorean theorem, the adjacent side (w) is equal to
hypotenuse a
a = x 2 + w2 ; a 2 = x 2 + w2 ; w2 = a 2 − x 2 ; w = a 2 − x 2
•
cos t =
adjacent
x
= . Therefore, the opposite side (w) is equal to
hypotenuse a
a = x 2 + w2 ; a 2 = x 2 + w2 ; w2 = a 2 − x 2 ; w = a 2 − x 2
•
tan t =
opposite x
= . Therefore, the hypotenuse (w) is equal to w = a 2 + x 2
adjacent a
•
cot t =
adjacent x
= . Therefore, the hypotenuse (w) is equal to w = a 2 + x 2
opposite a
Hamilton Education Guides
23
Advanced Integration
•
sec t =
1.2 Integration Using Trigonometric Substitution
hypotenuse x
= . Therefore, the opposite side (w) is equal to
adjacent
a
x = a 2 + w2 ; x 2 = a 2 + w2 ; w2 = x 2 − a 2 ; w = x 2 − a 2
•
csc t =
hypotenuse x
= . Therefore, the adjacent side (w) is equal to
opposite
a
x = a 2 + w2 ; x 2 = a 2 + w2 ; w2 = x 2 − a 2 ; w = x 2 − a 2
Let’s integrate some integrals using the above trigonometric substitution method:
Example 1.2-1: Use trigonometric substitution to evaluate the following indefinite integrals:
a. ∫
d. ∫
dx
=
b. ∫
=
2
e. ∫
x2 4 − x2
dx
(9 + x )
2
x 2 dx
=
c. ∫
dx =
f. ∫
25 − x 2
x2
2
x −1
1+ x 2
x2
dx =
dx
x
4
x 2 −1
=
Solutions:
a. Given ∫
(
dx
x
2
4− x
let x = 2 sin t , then dx = 2 cos t dt and 4 − x 2 = 4 − 4 sin 2 t = 4 1 − sin 2 t
2
)
= 4 cos 2 t = 2 cos t . Substituting these values back into the original integral we obtain:
2 cos t
2 cos t dt
dx
1
1
1
∫ x 2 4 − x 2 = ∫ (2 sin t )2 ⋅ 2 cos t = ∫ 4 sin 2 t ⋅ 2 cos t dt = ∫ 4 sin 2 t dt = 4 ∫ csc t dt = − 4 cot t + c
4− x 2
1 
2
+c = − 
x
4
2

1 cos t
1
+c = −
= −
4 sin t
4
2


1
4− x2 
4 − x 2 ⋅ 2/ 
− 
+
c
=
 + c

x
4
2/ ⋅ x



Check: To check the answer we start with the solution and find its derivative. The derivative
should match with the integrand, i.e., the algebraic expression inside the integral.
Note that not all the steps in finding the derivative is given. At this level, it is
expected that students are able to work through the details that are not shown (review
differentiation techniques described in Chapters 2 and 3 in Calculus 1 book).
Let y = −
1
4
4− x
x
2
−2 x
+ c , then y ′ = −
− x 2 − 4− x 2 ⋅ 4− x 2
1
= −
4
4− x 2
2
Hamilton Education Guides
x
(
1 2 4− x
4
− x 2 − 4− x 2
1
= −
4
4− x 2
2
x
)
= −
2
⋅ x − 1⋅ 4 − x 2
x2
2
− 4−1 x
1 2 4− x
+0 = −
4
x2
−2 x 2
2
1 − x2 − 4 + x2
1
1
−4
= −
=
4 x2 4 − x2
4 x2 4 − x2
x2 4 − x2
24
Advanced Integration
1.2 Integration Using Trigonometric Substitution
(
x 2 dx
b. Given ∫
25 − x
let x = 5 sin t , then dx = 5 cos t dt and 25 − x 2 = 25 − 25 sin 2 t = 25 1 − sin 2 t
2
)
= 25 cos 2 t = 5 cos t . Substituting these values back into the original integral we obtain:
x 2 dx
∫ 25 − x 2
25 sin 2 t ⋅ 5 cos t dt
=
5 cos t
= ∫
∫
1 − cos 2t
25 sin 2 t
dt = 25 sin 2 t dt = 25
dt
1
2
∫
∫

25  −1 x  25  x 25 − x
25
25
25  1

∫ ( 1 − cos 2t ) dt =  t − sin 2t  + c = t − sin t cos t + c =  sin  −  ⋅
=
25
2
=
25  −1 x  x
25  −1 x  25  x

 sin
−
25 − x 2  + c =
 sin
−

2 
5 2
2 
5  2  25

2 
Check: Let y =
2

2
5
2 
2
2 
5
2 5

+c


25 − x 2 + c
1 1
x
− 2x
25 − x 2 +
⋅ −
⋅ 

2 5 2
2 25 − x 2 2 

1− x
25
25
x x
sin −1 −
25 − x 2 + c , then y ′ =
2
2
5 2
1
25
=
25
2
 50 − 4 x 2 
25
25 − 2 x 2
 =
−
−

2 
2 25 − x 2 2 25 − x 2
 4 25 − x 
1+ x 2
c. Given ∫
x
2
)
(

 2 25 − x 2 − 2 x 2 
1  25 − x 2
x2
25
25
 =

 =
⋅ − 
−
−


2
2 5 
2 
2
2
2 25 − x 2
2 25 − x 
25 − x
2 25 − x
 4 25 − x


5
=
25 − 25 + 2 x 2
2 25 − x 2
=
1+ x 2 =
dx let x = tan t , then dx = sec 2 t dt and
2x 2
2 25 − x 2
1+ tan 2 t =
Substituting these values back into the original integral we obtain ∫
= ∫
(
)
sec t ⋅ 1 + tan 2 t dt
2
tan t
= ln sec t + tan t +
= ∫
1
cos t
∫ sin t
2
sec t + sec t tan 2 t
2
tan t
dt =
dt = ln sec t + tan t +
cos 2 t
∫
sec t tan 2 t
2
tan t
dt +
1+ x 2
x2
x2
=
25 − x 2
sec 2 t = sec t
dx =
sec t
Hamilton Education Guides
1 + x2 + x −
tan 2 t
sec t
∫ tan 2 t dt = ∫ sec t dt + ∫ tan 2 t dt
cos 2 t
cos t
∫ cos t sin 2 t dt = ln sec t + tan t + ∫ sin 2 t dt
= ln sec t + tan t + ∫ sin −2 t cos t dt = ln sec t + tan t + sin −1 t + c = ln sec t + tan t +
= ln sec t + tan t − csc t + c = ln
∫
sec t ⋅ sec 2 t dt
1
+c
sin t
1 + x2
+c
x
25
Advanced Integration
1.2 Integration Using Trigonometric Substitution
2
1+ x
1+ x 2 + x −
Check: Let y = ln
2
x
x
⋅ x − 1+ x 2
2


2 1+ x
1
2x
⋅
+ 1 −


x2
1+ x 2 + x  2 1+ x 2

+ c , then y ′ =
(
2 x + 1 + x 2   2
2 

  x −1− x 
⋅
−

2 
 2
1+ x 2 + x
2 1+ x 2
 x 1+ x 
)
 2x + 2 1 + x 2   2
2 

  x − 1+ x 
−
⋅
 =
 
1 + x 2 + x  2 1 + x 2   x 2 1 + x 2 
1
=
1
=
1+ x 2
+
dx
d. Given ∫
1
x 2 1+ x 2
x 2 1+ x 2 + 1+ x 2
=
(
x 2 1+ x 2
)
1
(
1+ x 2 1+ x 2
=
(
x 2 1+ x 2
)
) = 1+ x
2
x2
let x = 3 tan t , then dx = 3 sec 2 t dt and 9 + x 2 = 9 + (3 tan t )2 = 9 + 9 tan 2 t
(9 + x )
= 9(1 + tan t ) = 9 sec t Substituting these values back into the original integral we obtain
2 2
2
dx
2
∫ (9 + x2 )
=
= ∫
2
3 sec 2 t dt
3 sec 2 t dt
2
2
= ∫
2
( 9 sec t )
2
81 sec t sec t
=
1
27
dt
1
1 1
∫ sec 2 t = 27 ∫ cos t dt = 27 ⋅ 2 ∫ (1 + cos 2t ) dt
2


1 
x
3x 
1  1
1

+c
(t + sin t cos t ) + c = 1  tan −1 x + x ⋅ 3  + c =  tan −1 +
 t + sin 2t  + c =
54 
3 9 + x 2 
54  2
54
3
54
2
2

9
+
x
9
+
x


Check: Let
+
=
e. Given ∫
(
1 27 + 3 x 2 − 6 x 2
⋅
2
54
9 + x2
(
(
18
18 9 + x
x2
2
)
1
9
1
1
1 3 9 + x 2 − 2 x ⋅ 3x
1  −1 x
3x 
 + c then y ′ =
 tan
+ ⋅
⋅
= ⋅
+
54  x 2  54
54
54 
3 9 + x2 
2 2
31 + 9 
9+ x
3 9 + x2


x −1
)
)
2 2
=
=
1
⋅
54
3
(9 + x )
2
+
1 27 − 3 x 2
⋅
2
54
9 + x2
(
)
=
(
(
1
18 9 + x 2
)
+
(
)
9 − x2
(
18 9 + x 2
=
2
)
)
9 + x2 + 9 − x2
(
18 9 + x 2
)
2
1
(9 + x )
2 2
dx let x = sec t , then dx = sec t tan t and
x 2 −1 =
sec 2 t − 1 =
Substituting these values back into the original integral we obtain ∫
x2
x 2 −1
tan 2 t = tan t
dx =
∫
sec 2 t
sec t tan t dt
tan t
= ∫ sec 3 t dt = ∫ sec 2 t sec t dt . Let u = sec t and dv = sec 2 t , then du = sec t tan t and v = tan t .
Using the substitution formula uv − ∫ v du the integral ∫ sec 2 t sec t dt can be rewritten as
∫ sec t dt = sec t tan t − ∫ sec t tan t dt = sec t tan t − ∫ sec t ( sec t − 1) dt = sec t tan t − ∫ ( sec t − sec t ) dt
3
Hamilton Education Guides
2
2
3
26
Advanced Integration
1.2 Integration Using Trigonometric Substitution
= sec t tan t − ∫ sec 3 t dt + ∫ sec t dt
Note that ∫ sec 3 t dt = sec t tan t − ∫ sec 3 t dt + ∫ sec t dt therefore by moving − ∫ sec 3 t dt to the left
hand side of the equality we obtain
∫ sec t dt + ∫ sec t dt = sec t tan t + ∫ sec t dt thus 2∫ sec t dt = sec t tan t + ∫ sec t dt and ∫ sec t dt
3
3
3
(
1
sec t tan t + sec t dt
2
∫
3
) = 12 ( sec t tan t + ln sec t + tan t ) + c = 12  x x − 1 + ln x + x − 1  + c
2
1
2

Check: Let y =  x x 2 − 1 + ln x + x 2 − 1  + c , then y ′ =

2
1  2
2 x 2  1
1
x −1 +
+


2
2 x 2 −1  2 x + x 2 −1

2 x + x 2 − 1 
2
2


2
1
2
−
+
x
x
x
2
1  4 x 2 − 2
1
1
1


 =
=
× 1 +
+
+


2 x + x 2 −1
2
2  2 x 2 −1
2
2 x 2 −1
2 x 2 −1
 2 x −1 

(
=
f. Given ∫
4x 2
1 4x 2 − 2 + 2
=
=
2 2 x 2 −1
4 x 2 −1
dx
x
)
4
x2
x 2 −1
let x = sec t , then dx = sec t tan t dt and
2
x −1



2
x −1 
1
x 2 −1 =
sec 2 t − 1 =
tan 2 t = tan t
Substituting these values back into the original integral we obtain
∫ x 4 x 2 −1 = ∫ sec 4 t tan t = ∫ sec 3 t = ∫ cos t dt = ∫ cos t cos t dt = ∫ (1 − sin t ) cos t dt
sec t tan t dt
dx
(
)
= ∫ cos t − sin t cos t dt = ∫
2
 x 2 −1 
x −1
⋅
−
 x2 
x2


x 2 − x 2 +1
x
4
∫
3
2
2
x −1
Hamilton Education Guides
=
2
2
2
x − 1 1  x 2 − 1 
− 
+ c , then y ′ =

x
x
3



x 2 − x 2 +1
=
3
2
1
cos t dt − sin t cos t dt = sin t − sin 3 t + c =
3
2
Check: Let y =
=
dt
x 2 − x 2 +1
x 2 −1
x2
=
2 x2
2
x −1
− x 2 −1
x2
3


x2 −1 1  x2 −1 
− 
 + c
3
x
x


2
3  x 2 − 1  2
− 
 ⋅
x
3


2 x2
x 2 −1
− x 2 −1
x2
 x 2 −1 
1
x 2 −1
1
⋅
−
=
−
 2 
x 2 x 2 −1 x 4 x 2 −1
x 2 −1  x  x 2 x 2 −1
1
x2
1
x
4
x2 −1
27
Advanced Integration
1.2 Integration Using Trigonometric Substitution
Example 1.2-2: Use trigonometric substitution to evaluate the following indefinite integrals:
a. ∫
(
dx
b. ∫ x 2 4 − x 2 dx =
c. ∫
dx =
e. ∫ a 2 − x 2 dx =
f. ∫ x 2 − a 2 dx =
dx =
h. ∫ 9 − x 2 dx =
i. ∫
)
x2
d. ∫
4+ x
2
x2
g. ∫
9− x
2
dx
=
3
1− x 2 2
( 4 + 9x )
3
2 2
1
x
2
4 + x2
=
dx =
Solutions:
a. Given ∫
∫
(
dx
(
dx
)
3
1− x 2 2
= ∫
)
3
1− x 2 2
(
let x = sin t , then dx = cos t dt and 1 − x 2 = 1− sin 2 t = cos 2 t . Therefore,
cos t dt
= ∫
)
3
cos 2 t 2
Check: Let y =
cos t
3
cos t
dt =
1
2
1⋅ 1 − x 2 −
x
1− x
+ c , then y ′ =
2
x
sin t
∫ cos 2 t dt = ∫ sec t dt = tan t + c = cos t + c =
1− x
−2 x
2− 2 x 2 + 2 x 2
⋅x
2 1− x 2
2 1− x 2
=
2
+c
1− x2
1− x
=
2
2
(
2
)
3
1− x 2 2
=
1
(1 − x )
3
2 2
b. Given ∫ x 2 4 − x 2 dx let x = 2 sin t , then dx = 2 cos t dt and 4 − x 2 = 4 − (2 sin t )2 = 4 − 4 sin 2 t
(
4 1 − sin 2 t
∫x
2
) = 4 cos t = 2 cos t . Therefore,
2
4 − x 2 dx =
2
2
)
(
1
∫ 4 sin t ⋅ 2 cos t ⋅2 cos t dt = 16∫ sin t cos t dt = 16∫ 4 (1 − cos 2t ) (1 + cos 2t ) dt
2
1
2
= 4∫ 1 − cos 2 2t dt = ∫ 4 dt − 4∫ cos 2 2t dt = ∫ 4 dt − 4∫ ( 1 + cos 4t ) dt = ∫ 4 dt − ∫ 2 dt − 2∫ cos 4t dt
1
2
(
4
2
)
= 4t − 2t − sin 4t + c = 4t − 2t − sin 2t cos 2t + c = 2t − 2 (sin t cos t ) cos 2 t − sin 2 t + c
x
x
= 2 sin −1 − 2  ⋅
2
2

3

4 − x 2   4 − x 2 x 2 
−1 x
2  2x − x 
−
−
2
sin
4
x
+c
−
+
c
=
  4


2
4
2
4 



Check: Let y = 2 sin −1
=
2
4 − x2
−
 2x − x 3 
x
 + c , then y ′ =
− 4 − x2 
 4 
2


(
)(
− 2 x 2 + x 4 + 2 − 3x 2 4 − x 2
Hamilton Education Guides
4 4 − x2
)=
2
2
2 1 − x4
2
4 − x2
−
−
(
− 2x 2x − x 3
8 4 − x2
) + (2 − 3x )(4 − x )
2
1
2 2
− 2 x 2 + x 4 + 8 − 14 x 2 + 3 x 4
4 4 − x2
28
Advanced Integration
8 − 4 x 4 + 16 x 2 − 8
=
c. Given ∫
1.2 Integration Using Trigonometric Substitution
4 4− x
dx
( 4 + 9x )
2
dx
∫ 4
=
3
− 4 x 4 + 16 x 2
=
4 4− x
( +x )
2 2
(
4
9
∫
=
(
)
3
4 + 9x 2 2
1
12
= ∫
(
2 sec 2 t
3
)
3
4 sec 2 2
4− x
2
=
(
)
4 4 − x2 x2 4 − x2
4 (4 − x )
dt =
∫
d. Given ∫
x 2 dx
4+ x
(
= 4 1 + tan 2 t
x 2 dx
∫ 4 + x2
= ∫
2

2
2 sec 2 t
3
3
2× 3
4 2 sec 2 t
dt =
1
2 sec 2 t
3
2
dt
x
3x
+ c , then y ′ =
2
dt
∫ 2 4 3 sec 3 t dt = 3 ∫ 2 64 sec t = 3 ∫ 8 sec t
1⋅ 4 4 + 9 x − 4 ⋅
4 4 + 9x 2
2
) = 4 sec t . Therefore,
2
x
= x2 4 − x2
2
3
∫ sec t = 12 ∫ cos t dt = 12 sin t + c = 12 4 + 9 x 2 + c = 4 4 + 9 x 2 + c
Check: Let y =
2
2
3
1
1
dt
4 4− x
4 − x2
⋅
2
2
3
= 4 + 9 ⋅ tan 2 t = 4 + 4 tan 2 t = 4 1 + tan 2 t
dx
− 4 x 4 + 16 x 2
let x = tan t , then dx = sec 2 t dt and 4 + 9 x 2 = 4 + 9 ⋅  tan t 
3
2 2
9
2
=
(
18 x
2 4+9 x 2
16 4 + 9 x 2
)
)
(
4 4+9 x 2 −36 x 2
⋅x
=
(
4+9 x 2
16 4 + 9 x 2
)
=
1
( 4 + 9x )
3
2 2
let x = 2 tan t , then dx = 2 sec 2 t dt and 4 + x 2 = 4 + (2 tan t )2 = 4 + 4 tan 2 t
) = 4 sec t = 2 sec t . Therefore,
2
4 tan 2 t ⋅ 2 sec 2 t dt
= 4 tan 2 t sec t dt = 4
2 sec t
∫
∫ ( sec t − 1) sec t dt = 4∫ ( sec t − sec t ) dt
2
3
= 4∫ sec 3 t dt − 4∫ sec t dt . To solve ∫ sec 3 t dt = ∫ sec 2 t ⋅ sec t dt use substitution method by
letting u = sec t and dv = sec 2 t then du = sec t tan t dt and v = tan t . Using the substitution
formula uv − ∫ v du we obtain ∫ sec 2 t ⋅ sec t dt = sec t tan t − ∫ tan t ⋅ sec t tan t dt
(
)
(
)
= sec t tan t − ∫ sec t tan 2 t dt = sec t tan t − ∫ sec t sec 2 t − 1 dt = sec t tan t − ∫ sec 3 t − sec t dt
= sec t tan t − ∫ sec 3 t dt + ∫ sec t dt . Again at this point we know that ∫ sec 3 t dt
= sec t tan t − ∫ sec 3 t dt + ∫ sec t dt bringing − ∫ sec 3 t dt into the left hand side of the equation we
obtain ∫ sec 3 t dt + ∫ sec 3 t dt = sec t tan t + ∫ sec t dt . Therefore 2∫ sec 3 t dt = sec t tan t + ∫ sec t dt
thus ∫ sec 3 t dt =
[
1
sec t tan t + sec t dt
2
∫
] = 12 [ sec t tan t + ln sec t + tan t ] . Now substituting this
value into 4∫ sec 3 t dt − 4∫ sec t dt we have
Hamilton Education Guides
29
Advanced Integration
∫
1.2 Integration Using Trigonometric Substitution
∫
4 sec 3 t dt − 4 sec t dt = 4 ⋅
]− 4∫ sec t dt = 2 [ sec t tan t + ln sec t + tan t ]
1
sec t tan t + ln sec t + tan t
2
[
= 2 sec t tan t + 2 ln sec t + tan t − 4 ln sec t + tan t + c = 2 sec t tan t
− 4 ln sec t + tan t + c
4 + x2 x
⋅ − 2 ln
2
2
− 2 ln sec t + tan t + c = 2 ⋅
x 4+ x2
x + 4+ x2
4 + x2 x
+c
− 2 ln
+ +c =
2
2
2
2
x 4 + x2
x + 4 + x2
1
2 x 2 
− 2 ln
+ c , then y ′ =  4 + x 2 +

2
2
2
2 4 + x2 

Check: Let y =

2 


2 4 + x 2 + 2x
2
1 
2x
  + 0 = 1  4 + 2x  −
−2
⋅ ⋅ 1 +
 x + 4 + x2 2  2 4 + x2  
2  4 + x2  

  x + 4 + x2  4 + x2






(2 + x ) x + 4 + x  − 2 4 + x − 2 x (2 + x ) x + 4 + x  − 2 4 + x − 2 x x − 4 + x
2
=
2
=
=
2
⋅
 x + 4 + x2  4 + x2




x 4 − 4x 2 − x 4
− 4 4 + x2
=
− 4x 2
− 4 4 + x2
2
⋅
 x + 4 + x2  4 + x2




(2 + x ) x + 4 + x  − 2 4 + x − 2 x x − 4 + x
2
2
2
 x + 4 + x2  4 + x2




2
=
2
x− 4+ x
2
=
2
2
x − 4 + x2
(
x4 − x3 4 + x2 − x3 4 + x2 − x2 4 + x2
−4 4+ x
)
2
x2
=
4 + x2
e. Given ∫ a 2 − x 2 dx let x = a sin t , then dx = a cos t dt and a 2 − x 2 = a 2 − a 2 sin 2 t
(
= a 2 1− sin 2 t
) = a 2 cos 2 t = a cos t . Therefore,
∫
∫
a 2 cos 2 t dt =
a2
2
2
a  1

∫ ( 1 + cos 2t ) dt =  t + sin 2t  + c
∫
a 2 − x 2 dx =
=

 2
2 
2
2 
x a2 − x2 
a2
( t + sin t cos t ) + c = a  sin −1 x + x ⋅ a − x  + c =  a sin −1 x +
 + c
2
a
2 
2
a a
a
 2





Check: Let y =
=
=
a cos t ⋅ a cos t dt =
2 
a2
a2
x x a2 − x2
sin −1 +
+ c , then y ′ =
2
a
2
2
a2 − x2 − x2
a2
a
⋅
+
2 a a2 − x2
2 a2 − x2
a2 − x2
2
a −x
Hamilton Education Guides
2
⋅
a2 − x2
2
a −x
2
=
a2
2 a2 − x2
=
+
1
( )2
⋅
1 − ax
a 2 − 2x 2
2 a2 − x2
=
2
2
2
2
a −x

1 1
− 2x 2
+
a2 − x2 +
a 2
2 a2 − x2
2a 2 − 2 x 2
2 a2 − x2
(a − x ) a − x
2
2
2
=
=
a2 − x2
a2 − x2
a2 − x2
30
Advanced Integration
1.2 Integration Using Trigonometric Substitution
x2 − a2 =
f. Given ∫ x 2 − a 2 dx let x = a sec t , then dx = a sec t tan tdt and
a 2 tan 2 t
=
a 2 sec 2 t − a 2
= a tan t . Therefore,
∫ x − a dx = ∫ a tan t ⋅ a sec t tan t dt = ∫ a sec t tan t dt = a ∫ sec t ( sec t − 1) dt = a ∫ sec t dt
2
2
2
∫
=

x
a2  x x2 − a2
− ln +
⋅

a
a
2 a

=
x
2
=
g. Given ∫
3
2
2

x
a2
x + x2 − a2

2
2
x
−
a
−
+c
ln
c
+
=

a
2
2


x2 − a2
a
x
a2
a2
ln a + c =
ln x + x 2 − a 2 +
2
2
2
Check: Let y =
−
2
2
a2
(tan t sec t + ln sec t + tan t ) − a 2 ln sec t + tan t + c = a (tan t sec t − ln sec t + tan t ) + c
2
2
− a 2 sec t dt =
x2 − a2 −
2
x
2
x2 − a2 −
x2 −a2 −
a2
ln x + x 2 − a 2 + c
2

1
2x 2
a2

ln x + x 2 − a 2 + c , then y ′ =  x 2 − a 2 +
2
2
2
2 
2 x −a 



a2
1
x2 − a2 + x2 a2
x + x2 − a2
1
2x
+0 =
⋅
⋅ 1 +
−
⋅
⋅
2 x + x 2 − a 2  2 x 2 −a 2 
2 x + x2 − a2
2 x2 − a2
x2 − a2
x2 − a2 + x2
2
2 x −a
x2
9− x
2
2
−
a2
2
2 x −a
2
=
2 x 2 − 2a 2
2
2 x −a
2
=
x2 − a2
2
x −a
2
x2 − a2
=
2
x −a
9 − x2 =
dx let x = 3 sin t , then dx = 3 cos t dt and
2
⋅
x2 − a2
2
x −a
2
9 − 9 sin 2 t =
x2 − a2
=
(
9 1 − sin 2 t
)
= 9 cos 2 t = 3 cos t . Substituting these values back into the original integral we obtain:
x 2 dx
∫ 9 − x2
= ∫
9 sin 2 t ⋅ 3 cos t dt
3 cos t
= ∫
1 − cos 2t
9 sin 2 t
dt = 9 sin 2 t dt = 9
dt
1
2
∫
∫

2 
9  −1 x  9  x 9 − x 
9
9
9 1

+c
∫ ( 1 − cos 2t ) dt =  t − sin 2t  + c = t − sin t cos t + c =  sin  −  ⋅
=
9
2
=
x x
9
9  −1 x  9  x

9− x2 + c
9 − x 2  + c =  sin −1  −
− 
 sin
2
3 2
3 29
2

Check:
2
Let
y=
2

2
x x
9
sin −1 −
9 − x2 + c ,
2
3 2
2
2
then
y′
=
9
2
3 23

1 1
⋅ −
2 3 2

1− x
1
9
Hamilton Education Guides
3
9 − x2 +


x
⋅ 

2 9 − x2 2 
− 2x
31
Advanced Integration

1 
⋅ −
9 − x 2 3 
9
2
=
1.2 Integration Using Trigonometric Substitution
9 − x2
3
−
2
(


9
9 − 2x 2
 18 − 4 x 2 
=
−
−

 4 9 − x2 
2 9 − x2 2 9 − x2


=
9 − 9 + 2x 2
9 cos 2 t = 9 cos t
2x 2
=
2 9 − x2
h. Given ∫ 9 − x 2 dx let x = 3 sin t , then dx = 3 cos t dt and
=
)


2
2 
9
9

 2 9 − x − 2x 
=
−


 =
2 
2
2
 4 9− x

2 9 − x2
2 9− x 
2 9− x


x2
2 9 − x2
9 − x2
x2
=
9 − x2
(
9 1 − sin 2 t
9 − 9 sin 2 t =
=
)
. Therefore,
9
1
9

∫ 9 − x dx = ∫ 3 cos t ⋅ 3 cos t dt = ∫ 9 cos t dt = 2 ∫ ( 1 + cos 2t ) dt = 2  t + 2 sin 2t  + c
2
2
=

9
( t + sin t cos t ) + c = 9  sin −1 x + x ⋅
2
2
3 3

9
2
x
3
Check: Let y = sin −1 +
9
⋅
2 3
=
9− x
9 − x2
=
i. Given ∫
3
9− x
2
1
+
2
2
9 − x2 − x2
2
9− x
9 − x2
=
2
9 − x2
x
9− x
2
2

x
9 − x 2 
9
−1 x
+
c
=
 2 sin 3 +

3





9− x2 
+c
2


9
2
9 − x2 +
+ c , then y ′ =
=
9
2
9 − x2
⋅
9− x
2
9− x
=
2
+
1
1 1
⋅ +
3 2
2
x
( )
1− 3
9 − 2x 2
2
9− x
=
2
2
9− x
(9 − x ) 9 − x = 9 − x
2
2
9− x
2
=
2
2
9 − x2
18 − 2 x 2
2
9− x
2
=
(
2 9 − x2
2
9− x
)
2
2
4 + x2 =
dx let x = 2 tan t , then dx = 2 sec 2 t dt and
x2 4 + x2
9 + 9 − 2x 2
− 2x 2
4 + 4 tan 2 t = 2 1 + tan 2 t
= 2 sec 2 t = 2 sec t . Therefore,
dx
∫ x2 4 + x2
=
1
4
cos t
∫ u2
⋅
2 sec 2 t
dx =
∫ 4 tan 2 t ⋅ ( 2 sec t )
du
cos t
1
4
1
∫ u2
du = −
1
4
4+ x
x
2
2 
2 
=
Check: Let y = −
dt =
1
4
Hamilton Education Guides
4+ x 2
x
2
sec dt
∫ tan 2 t
dt =
1
4
∫
1 cos 2 t
1
⋅
dt =
2
cos t sin t
4
1
1 −1
1
1
u +c = −
+c = −
+c = −
4
4
4u
4 sin t
2x
+ c , then y ′ = −
2
x − 4 + x ⋅ 4 + x 



= −
1
4
1 2 4+ x
4
2
x
4+ x2
+c
x
2 x2
⋅ x − 1⋅ 4 + x 2
= −
2
cos t
∫ sin 2 t dt let u = sin t
2
− 4 + x2
1 2 4+ x
4
x2
x 2 − 4− x 2
= −
4+ x 2
4x
2
= −
−4
4x
2
4+ x
2
=
1
x
2
4 + x2
32
Advanced Integration
1.2 Integration Using Trigonometric Substitution
The following are additional standard forms of integration that have already been derived.
Trigonometric substitution can be used in most of these cases in order to confirm the result.
Table 1.2-1: Integration Formulas
dx
1
−1 x
1.
∫ a 2 + x2 = a tan a + c
3.
∫ a 2 − x 2 = 2a ln x − a + c = 2a ln a − x + c
5.
∫ a 2 + x 2 = sinh
7.
∫x
dx
1
dx
∫
9.
−1 x
(
x a 2 + 2x 2
8
) a + x − a sinh x + c
2
2
4
−1
8
a
1
x
4.
∫ ( a 2 − x 2 ) 2 = 2a 2 ( a 2 − x 2 ) + 2a 2 ∫ a 2 − x 2
6.
∫ a + x dx = 2 a + x + 2 sinh
8.
∫
dx
2
1
x
x
2
2
a2 + x2
dx =
x
x2
a2
2
a 2 + x 2 − a sinh −1
−1 x
a
12.
∫ x2 a2 + x2 = −
+c
14.
∫
a 2 − x 2 dx =
16.
∫
a2 − x2
dx =
x
x
a2 − x2
−
+c
a
x
18.
∫ a 2 − x 2 dx = 2 sin
a + a2 − x2
+c
x
20.
∫ x2 a2 − x2 =
+ c = ln x + x 2 − a 2 + c
22.
∫ x − a dx = 2 x − a − 2 cosh
dx
−1 x
13.
∫ a 2 − x 2 = sin
15.
∫
a
(
∫
a2 − x2
x
2
dx
dx = − sin −1
1
a
19.
∫ x a2 − x2
21.
∫ x 2 − a 2 = cosh
−1 x
23.
∫x
x
2x 2 − a 2
8
= − ln
dx
2
x 2 − a 2 dx =
x2 − a2
a
(
a
a2 − x2
dx
a2x
x
2
−1 x
−
+c
a2
2
2
1
x a2 − x2 + c
2
−1 x
a
+c
∫ x 2 − a 2 dx = 2 cosh
x
1
a
+ c = cos −1
+c
a
x
a
28.
∫ x2 x2 − a2 =
30.
∫ a + x dx = 2 a + x + 2 ln  x + x + a  + c
29.
∫ 2ax − x 2 = sin  a  + c
dx
a2
a + a2 − x2
+c
x
26.
x
−
a
∫ x x2 − a2 = a sec
−1
a 2 − x 2 − a ln
x2 − a2
+c
x
dx = cosh −1
1
a2
x
x
a2 − x2 +
sin −1 + c
a
2
2
x2
2
+c
a2x
∫
27.
dx
a2 + x2
dx
24.
∫
x
dx = −
) x − a − a8 cosh ax + c
25.
2
)
a4
x 1
sin −1 − x a 2 − x 2 a 2 − 2 x 2 + c
8
a 8
x 2 a 2 − x 2 dx =
+c
a2
x x a2 + x2
+c
sinh −1 +
2
2
a
a + a2 + x2
+c
x
1
+c
a
+c
x
∫ a2 + x2
dx
a
dx
10.
∫ x a 2 + x 2 = − a ln
−1 x
∫ ( a 2 + x 2 )2 = 2a 2 ( a 2 + x 2 ) + 2a 3 tan
a2 + x2
x
a2 + x2
+c
dx = sinh −1 −
a
x
x
11.
17.
a+x
+ c = ln x + a 2 + x 2 + c
a
a 2 + x 2 dx =
2
1
x+a
dx
2.
−1  x − a 
Hamilton Education Guides
2
2
4
−1
x2 − a2
dx =
x
x2
a2
dx
2
x 2 − a 2 − a sec −1
2
x
−1 x
a
x2 − a2
a2x
2
2
+
x
2
x
+c
a
x2 − a2 + c
+c
a2

2
2
33
Advanced Integration
1.2 Integration Using Trigonometric Substitution
Section 1.2 Practice Problems – Integration Using Trigonometric Substitution
Evaluate the following indefinite integrals:
a. ∫
dx
x
2
16 − x
1
2
=
x2
b. ∫
9− x




+ 5 x  dx =
 2

 x −1

dx =
2
e. ∫ 
g. ∫ 36 − x 2 dx =
h. ∫
d. ∫
(49 + x )
2
Hamilton Education Guides
2
dx =
x2
dx
( 9 + 36 x )
3
2 2
=
c. ∫
dx
x 9 + 4x
2
=
f. ∫ x 2 − 25 dx =
i. ∫
9 − 4x 2
dx =
x
34
Advanced Integration
1.3
1.3 Integration by Partial Fractions
Integration by Partial Fractions
The primary objective of this section is to show that rational functions can be integrated by
breaking them into simpler parts. A function F (x ) =
f (x )
, where f (x ) and g (x ) are polynomials,
g (x )
is referred to as a rational function. Depending on the degree of the f (x ) , the function F (x ) is
either a proper or an improper rational fraction.
•
A proper rational fraction is a fraction where the degree of the numerator f (x ) is less than
the degree of the denominator g (x ) . For example,
are proper rational fractions.
•
x +1
2
x − x−6
,
1
3
x −1
,
1
2
x + 2x − 3
, and
x
2
x +1
An improper rational fraction is a fraction where the degree of the numerator f (x ) is at least
as large as the degree of the denominator g (x ) . In such cases, long division is used in order
to reduce the fraction to the sum of a polynomial and a proper rational fraction. For example,
x3 + 2
x2 − x − 6
,
x3
x 2 +1
, and
x 4 − x 3 − x −1
x3 − x2
are improper rational fractions. (See Section 6.3 of
Mastering Algebra – An Introduction for solved problems on long division.)
A fraction, depending on its classification of the denominator, can be represented in four
different cases. These cases are as follows:
CASE I - The Denominator Has Distinct Linear Factors
In this case the linear factors of the form ax + b appear only once in the denominator. To solve
this class of rational fractions we equate each proper rational fraction with a single fraction of the
form
C
A
B
,
,
, etc.
ax + b cx + d ex + f
The following examples show the steps as to how this class of
integrals are solved.
Example 1.3-1: Evaluate the integral ∫
x +1
3
x + x 2 − 6x
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second - Factor the denominator x 3 + x 2 − 6 x into x(x − 2)(x + 3) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
x +1
3
2
x + x − 6x
=
x +1
A
B
C
= +
+
x(x − 2 )(x + 3)
x x−2 x+3
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x +1
x 3 + x 2 − 6x
Hamilton Education Guides
=
A (x − 2 )(x + 3) + Bx (x + 3) + Cx (x − 2 )
x(x − 2 )(x + 3)
35
Advanced Integration
1.3 Integration by Partial Fractions
) (
) (
(
x + 1 = A x 2 + 3x − 2 x − 6 + B x 2 + 3x + C x 2 − 2 x
) = Ax + Ax − 6 A + Bx + 3Bx + Cx − 2Cx
2
2
x + 1 = ( A + B + C )x 2 + ( A + 3B − 2C )x − 6 A
A+ B +C = 0
2
therefore,
A + 3B − 2C = 1
1
6
which result in having A = − , B =
−6 A = 1
3
2
, and C = −
10
15
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x +1
A
B
1
C
1
3
1
2
1
∫ x 3 + x 2 − 6 x dx = ∫ x dx + ∫ x − 2 dx + ∫ x + 3 dx = − 6 ∫ x dx + 10 ∫ x − 2 dx − 15 ∫ x + 3 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
−
1
6
1
3
2
1
1
1
3
2
∫ x dx + 10 ∫ x − 2 dx − 15 ∫ x + 3 dx = − 6 ln x + 10 ln x − 2 − 15 ln x + 3 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
1
6
Let y = − ln x +
=
=
=
1
2
1
3
1 1
2
3
⋅1 + 0
⋅1 − ⋅
ln x − 2 − ln x + 3 + c , then y ′ = − ⋅ ⋅1 + ⋅
15 x + 3
10 x − 2
6 x
15
10
(
)
(
)
)
(
− 150 x 2 + x − 6 + 270 x 2 + 3 x − 120 x 2 − 2 x
− 150(x − 2 )(x + 3) + 270 x (x + 3) − 120 x (x − 2 )
=
900 x (x − 2 )(x + 3)
900 x x 3 + x 2 − 6 x
− 150 x 2 − 150 x + 900 + 270 x 2 + 810 x − 120 x 2 + 240 x
(
900 x 3 + x 2 − 6 x
900 x + 900
(
900 x 3 + x 2 − 6 x
)
=
(
=
)
900(x + 1)
900 x 3 + x 2 − 6 x
=
)
Example 1.3-2: Evaluate the integral ∫
(
)
(− 150 + 270 − 120)x 2 + (− 150 + 810 + 240)x + 900
(
900 x 3 + x 2 − 6 x
)
x +1
3
x + x 2 − 6x
dx
2
x + 3x + 2
.
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second - Factor the denominator x 2 + 3x + 2 into (x + 1)(x + 2) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
1
2
x + 3x + 2
=
1
(x + 1)(x + 2)
=
A
B
+
x +1 x + 2
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
1
2
x + 3x + 2
Hamilton Education Guides
=
A (x + 2 ) + B (x + 1)
(x + 1)(x + 2)
36
Advanced Integration
1.3 Integration by Partial Fractions
1 = A (x + 2 ) + B (x + 1) = Ax + 2 A + Bx + B
1 = ( A + B )x + (2 A + B ) therefore,
2A + B = 1
A+ B = 0
which result in having A = 1 and B = −1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
A
1
B
1
∫ x 2 + 3x + 2 dx = ∫ x + 1 dx + ∫ x + 2 dx = ∫ x + 1 dx − ∫ x + 2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
∫ x + 1 dx − ∫ x + 2 dx = ln x + 1 − ln x + 2 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
(x + 2) − (x + 1) =
1
1
1
⋅1 −
⋅1 + 0 =
2
(x + 1)(x + 2)
x +1
x+2
x + 3x + 2
Let y = ln x + 1 − ln x + 2 + c , then y ′ =
Example 1.3-3: Evaluate the integral ∫
x dx
2
x − 5x + 6
.
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second - Factor the denominator x 2 − 5 x + 6 into (x − 2)(x − 3) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
x
2
x − 5x + 6
=
x
(x − 2)(x − 3)
=
A
B
+
x−2 x−3
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
x
2
x − 5x + 6
=
A (x − 3) + B (x − 2 )
(x − 2)(x − 3)
x = A (x − 3) + B (x − 2 ) = Ax − 3 A + Bx − 2 B
x = ( A + B )x − (3 A + 2 B )
A+ B =1
therefore,
3 A + 2B = 0
which result in having A = −2 and B = 3
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
Hamilton Education Guides
37
Advanced Integration
1.3 Integration by Partial Fractions
x dx
A
2
B
3
∫ x 2 − 5x + 6 = ∫ x − 2 dx + ∫ x − 3 dx = − ∫ x − 2 dx + ∫ x − 3 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
−
2
3
∫ x − 2 dx + ∫ x − 3 dx = − 2 ln x − 2 + 3 ln x − 3 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = −2 ln x − 2 + 3 ln x − 3 + c , then y ′ = − 2 ⋅
=
−2 x + 6 + 3 x − 6
2
x − 5x + 6
=
− 2(x − 3) + 3(x − 2 )
1
1
⋅1 + 3 ⋅
⋅1 + 0 =
(x − 2)(x − 3)
x−2
x−3
x
2
x − 5x + 6
Example 1.3-4: Evaluate the integral ∫
x 2 +1
x3 − x
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 − x into x x 2 − 1 = x(x − 1)(x + 1) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
x 2 +1
=
x3 − x
x 2 +1
A
B
C
= +
+
x(x − 1)(x + 1)
x x −1 x +1
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x 2 +1
3
x −x
(
=
A (x − 1)(x + 1) + Bx (x + 1) + Cx (x − 1)
x(x − 1)(x + 1)
) (
) (
x 2 +1 = A x 2 + x − x −1 + B x 2 + x + C x 2 − x
) = Ax − A + Bx + Bx + Cx − Cx
2
x 2 + 1 = ( A + B + C )x 2 + (B − C )x − A
A+ B +C =1
2
2
therefore,
−A = 1
B −C = 0
which result in having A = −1 , B = 1 , and C = 1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x 2 +1
A
B
C
1
1
1
∫ x 3 − x dx = ∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx = − ∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
−
1
1
1
∫ x dx + ∫ x − 1 dx + ∫ x + 1 dx = − ln x + ln x − 1 + ln x + 1 + c
Hamilton Education Guides
38
Advanced Integration
1.3 Integration by Partial Fractions
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
1
x
Let y = − ln x + ln x − 1 + ln x + 1 + c , then y ′ = − +
=
− x 2 +1+ x 2 + x + x 2 − x
(
)
x x 2 −1
− (x − 1)(x + 1) + x (x + 1) + x (x − 1)
1
1
+
+0 =
x(x − 1)(x + 1)
x −1 x +1
x 2 +1
=
x3 − x
Example 1.3-5: Evaluate the integral ∫
(
x−3
x x2 + x − 2
dx .
)
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second - Factor the denominator x 2 + x − 2 into (x + 2)(x − 1) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
x−3
A
B
C
= +
+
x(x + 2 )(x − 1)
x x + 2 x −1
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
A (x + 2 )(x − 1) + Bx (x − 1) + Cx (x + 2)
x−3
=
x(x + 2)(x − 1)
x(x + 2 )(x − 1)
(
) (
) (
x − 3 = A x 2 + 2x − x − 2 + B x 2 − x + C x 2 + 2x
) = Ax + Ax − 2 A + Bx − Bx + Cx + 2Cx
2
2
x − 3 = ( A + B + C )x 2 + ( A − B + 2C )x − 2 A
A+ B +C = 0
2
therefore,
−2 A = −3
A − B + 2C = 1
3
2
5
6
which result in having A = , B = − , and C = −
2
3
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x−3
A
B
3
C
1
5
1
2
1
∫ x (x 2 + x − 2) dx = ∫ x dx + ∫ x + 2 dx + ∫ x − 1 dx = 2 ∫ x dx − 6 ∫ x + 2 dx − 3 ∫ x − 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
3
2
1
5
1
2
1
3
5
2
∫ x dx − 6 ∫ x + 2 dx − 3 ∫ x − 1 dx = 2 ln x − 6 ln x + 2 − 3 ln x − 1 + c
Seventh - Check the answer by differentiating the solution. The result should match the
3
2
5
6
2
3
integrand.Let y = ln x − ln x + 2 − ln x − 1 + c , then y ′ =
(
) (
(
) (
)
5 1
2 1
3 1
⋅1 + 0 =
⋅ ⋅1 − ⋅
⋅1 − ⋅
3 x −1
6 x+2
2 x
9 x 2 + x − 2 − 5 x 2 − x − 4 x 2 + 2x
9(x + 2 )(x − 1) − 5 x (x − 1) − 4 x (x + 2 )
=
6 x (x + 2 )(x − 1)
6x x 2 + x − 2
Hamilton Education Guides
) = (9 − 5 − 4)x + (9 + 5 − 8)x − 18
6 x (x + x − 2 )
2
2
39
Advanced Integration
=
(
6 x − 18
6x x 2 + x − 2
)
1.3 Integration by Partial Fractions
=
(
6(x − 3)
6x x 2 + x − 2
)
=
x−3
(
x x2 + x − 2
Example 1.3-6: Evaluate the integral ∫
x3 + 2
x2 − x − 6
)
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction. In this case the
integrand is an improper rational fraction, i.e., the degree of the numerator is greater than the
degree of the denominator. Applying the long division method we obtain
x3 + 2
2
x − x−6
= (x + 1) +
7x + 8
2
x − x−6
To integrate the second term we proceed with the following steps:
Second - Factor the denominator x 2 − x − 6 into (x − 3)(x + 2) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
7x + 8
2
x − x−6
=
7x + 8
(x − 3)(x + 2)
=
A
B
+
x−3 x+ 2
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
7x + 8
2
x − x−6
=
A ( x + 2 ) + B ( x − 3)
(x − 3)(x + 2)
7 x + 8 = Ax + 2 A + Bx − 3B
7 x + 8 = ( A + B )x + (2 A − 3B )
2 A − 3B = 8
A+ B = 7
which result in having A =
therefore,
6
29
and B =
5
5
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x3 + 2
7x + 8
7x + 8
1
A
B
∫ x 2 − x − 6 dx = ∫ (x + 1)dx + ∫ x 2 − x − 6 dx = ∫ (x + 1)dx + ∫ (x − 3)(x + 2) dx = 2 (x + 1) + ∫ x − 3 dx + ∫ x + 2 dx
=
1
(x + 1)2 + 29
2
5
1
6
2
1
∫ x − 3 dx + 5 ∫ x + 2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
(x + 1)2 + 29
2
5
1
6
1
1
29
6
∫ x − 3 dx + 5 ∫ x + 2 dx = 2 (x + 1) + 5 ln x − 3 + 5 ln x + 2 + c
2
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Hamilton Education Guides
40
Advanced Integration
1
2
Let y = (x + 1)2 +
= (x + 1) +
29
6
6 1
29 1
ln x − 3 + ln x + 2 + c , then y ′ = (x + 1) +
⋅1 + 0
⋅1 + ⋅
⋅
5
5
5 x+2
5 x−3
5(x + 1)(x − 3)(x + 2 ) + 29(x + 2 ) + 6(x − 3)
5 x 3 − 35 x − 30 + 29 x + 58 + 6 x − 18
6 1
29 1
=
=
+ ⋅
⋅
5 (x − 3)(x + 2 )
5 (x − 3)(x + 2 )
5 x−3 5 x+ 2
5 x 3 + 10
=
5 (x − 3)(x + 2)
=
1.3 Integration by Partial Fractions
(
(
5 x3 + 2
)
5 x2 − x − 6
)
=
x3 + 2
x2 − x − 6
Example 1.3-7: Evaluate the integral ∫
1
49 − x 2
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second - Factor the denominator 49 − x 2 into (7 − x )(7 + x ) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the
denominator is occurring only once, the integrand can be represented in the following way:
1
49 − x
2
=
1
(7 − x )(7 + x )
=
A
B
+
7−x 7+ x
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
1
49 − x
=
2
A (7 + x ) + B (7 − x )
(7 − x )(7 + x )
1 = A (7 + x ) + B (7 − x ) = 7 A + Ax + 7 B − Bx
1 = ( A − B )x + (7 A + 7 B ) therefore,
7 A + 7B = 1
which result in having A =
A− B = 0
1
1
, and B =
14
14
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
A
1
B
1
1
1
∫ 49 − x 2 dx = ∫ 7 − x dx + ∫ 7 + x dx = 14 ∫ 7 − x dx + 14 ∫ 7 + x dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
14
1
1
1
1
1
∫ 7 − x dx + 14 ∫ 7 + x dx = 14 ln 7 − x + 14 ln 7 + x + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
(
1
1
7+ x+7−x
1
1
1
1
ln 7 − x + ln 7 + x + c , then y ′ =
⋅
+ ⋅
+0 =
14
14
14(7 − x )(7 + x )
14 7 − x 14 7 + x
7+7
14 49 + 7 x − 7 x − x 2
)
Hamilton Education Guides
=
(
14
14 49 − x 2
)
=
1
49 − x 2
41
Advanced Integration
1.3 Integration by Partial Fractions
CASE II - The Denominator Has Repeated Linear Factors
In this case each linear factor of the form ax + b appears n times in the denominator. To solve
this class of rational fractions we equate each proper rational fraction, that appears n times in the
denominator, with a sum of n partial fractions of the form
Mn
M1
M2
+
+ ... +
. The
2
ax + b (ax + b )
(ax + b )n
following examples show the steps as to how this class of integrals are solved.
Example 1.3-8: Evaluate the integral ∫
x+3
3
x − 2x 2 + x
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 − 2 x 2 + x into x x 2 − 2 x + 1 = x(x − 1)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the
denominator is repeated, the integrand can be represented in the following way:
x+3
3
2
x − 2x + x
=
(
x+3
=
)
x x 2 − 2x +1
x+3
x(x − 1)
2
=
A
B
C
+
+
x x − 1 (x − 1)2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x+3
x 3 − 2x 2 + x
(
) (
=
A (x − 1)2 + Bx (x − 1) + Cx
x(x − 1)(x − 1)2
)
x + 3 = A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx
x + 3 = ( A + B )x 2 + (− 2 A − B + C )x + A
A+ B = 0
therefore,
A=3
−2 A − B + C = 1
which result in having A = 3 , B = −3 , and C = 4
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
A
x+3
B
C
1
1
1
∫ x 3 − 2 x 2 + x dx = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = 3∫ x dx − 3∫ x − 1 dx + 4∫ (x − 1)2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
1
4
∫ x dx − 3∫ x − 1 dx + 4∫ (x − 1)2 dx = 3 ln x − 3 ln x − 1 − x − 1 + c
3
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = 3 ln x − 3 ln x − 1 −
Hamilton Education Guides
3(x − 1)2 − 3 x (x − 1) + 4 x
1
1
4
4
+
+0 =
+ c , then y ′ = 3 ⋅ − 3 ⋅
x
x − 1 (x − 1)2
x −1
x (x − 1)2
42
Advanced Integration
=
1.3 Integration by Partial Fractions
3x 2 − 6 x + 3 − 3x 2 + 3x + 4 x
3
x+3
=
2
x − 2x + x
3
x − 2x 2 + x
Example 1.3-9: Evaluate the integral ∫
dx
3
x − x2
.
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second - Factor the denominator x 3 − x 2 into x 2 (x − 1) .
Third - Write the linear factors in partial fraction form. Since one of the factors in the
denominator is repeated, the integrand can be represented in the following way:
1
3
x −x
2
=
1
x (x − 1)
2
A B
C
+
+
2
x x
x −1
=
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
x3 − x
(
=
2
Ax (x − 1) + B (x − 1) + Cx 2
x 2 (x − 1)
)
1 = A x 2 − x + B (x − 1) + Cx 2 = Ax 2 − Ax + Bx − B + Cx 2
1 = ( A + C )x 2 + (− A + B )x − B therefore,
A+C = 0
−B = 1
−A + B = 0
which result in having A = −1 , B = −1 , and C = 1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
dx
A
B
1
C
1
1
∫ x 3 − x 2 = ∫ x dx + ∫ x 2 dx + ∫ x − 1 dx = − ∫ x dx − ∫ x 2 dx + ∫ x − 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
−
1
1
1
1
1
∫ x dx − ∫ x 2 dx + ∫ x − 1 dx = − ln x + x + ln x − 1 + c = x + ln x − 1 − ln x + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
1
x
Let y = + ln x − 1 − ln x + c , then y ′ = −
=
− x +1+ x 2 − x 2 + x
3
x −x
2
=
x2
+
− (x − 1) + x 2 − x (x − 1)
1
1
− +0 =
x −1 x
x 2 (x − 1)
1
3
x − x2
Example 1.3-10: Evaluate the integral ∫
Hamilton Education Guides
1
5dx
3
x − 2x 2 + x
.
43
Advanced Integration
1.3 Integration by Partial Fractions
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Second – Factor the denominator x 3 − 2 x 2 + x into x(x − 1)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the
denominator is repeated, the integrand can be represented in the following way:
5
x(x − 1)
A
B
C
+
+
x x − 1 (x − 1)2
=
2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
5
x 3 − 2x 2 + x
(
) (
A (x − 1)2 + Bx (x − 1) + Cx
=
x(x − 1)2
)
5 = A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx
5 = ( A + B )x 2 + (− 2 A − B + C )x + A
A+ B = 0
therefore,
−2 A − B + C = 0
A=5
which result in having A = 5 , B = −5 , and C = 5
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
A
5dx
B
1
C
1
1
∫ x 3 − 2 x 2 + x = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = 5∫ x dx − 5∫ x − 1 dx + 5∫ (x − 1)2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
5
1
1
1
5
∫ x dx − 5∫ x − 1 dx + 5∫ (x − 1)2 dx = 5 ln x − 5 ln x − 1 − x − 1 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = 5 ln x − 5 ln x − 1 −
=
5(x − 1)2 − 5 x (x − 1) + 5 x
x(x − 1)
2
=
1
1
1
5
5
5
5
⋅1 + 0 = −
⋅1 + 5 ⋅
+
+ c , then y ′ = 5 ⋅ ⋅1 − 5 ⋅
2
1
x −1
x
−
x
x
x −1
(x − 1)
(x − 1)2
(
)
5 x 2 − 2 x + 1 − 5x 2 + 5x + 5x
3
2
x − 2x + x
Example 1.3-11: Evaluate the integral ∫
x+6
(x + 2)(x − 3)2
=
5 x 2 − 10 x + 5 − 5 x 2 + 5 x + 5 x
3
2
x − 2x + x
=
5
3
x − 2x 2 + x
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Hamilton Education Guides
44
Advanced Integration
1.3 Integration by Partial Fractions
Second – Factor the denominator. However, the denominator is already in its reduced form of
(x + 2)(x − 3)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the
denominator is repeated, the integrand can be represented in the following way:
x+6
(x + 2)(x − 3)
=
2
A
B
C
+
+
x + 2 x − 3 (x − 3)2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x+6
(x + 2)(x − 3)
) (
(
A (x − 3)2 + B(x + 2)(x − 3) + C (x + 2)
=
2
(x + 2)(x − 3)2
)
x + 6 = A x 2 − 6 x + 9 + B x 2 − x − 6 + C (x + 2 ) = Ax 2 − 6 Ax + 9 A + Bx 2 − Bx − 6 B + Cx + 2C
x + 6 = ( A + B )x 2 + (− 6 A − B + C )x + (9 A − 6 B + 2C )
A+ B = 0
therefore,
9 A − 6 B + 2C = 6
−6 A − B + C = 1
which result in having A =
9
4
4
, B = − , and C =
5
25
25
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x+6
A
B
4
C
1
4
1
9
1
∫ (x + 2)(x − 3)2 dx = ∫ x + 2 dx + ∫ x − 3 dx + ∫ (x − 3)2 dx = 25 ∫ x + 2 dx − 25 ∫ x − 3 dx + 5 ∫ (x − 3)2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
4
25
1
4
1
9
4
1
4
9
1
∫ x + 2 dx − 25 ∫ x − 3 dx + 5 ∫ (x − 3)2 dx = 25 ln x + 2 − 25 ln x − 3 − 5 (x − 3) + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
+
4
4
1
4
1
9
1
4
9 1
⋅
− ⋅
+
+0
ln x + 2 − ln x − 3 −
+ c , then y ′ =
25 x + 2 25 x − 3 5 (x − 3)2
25
25
5 (x − 3)
4(x − 3)2 − 4(x + 2 )(x − 3) + 45(x + 2 )
25(x + 2 )(x − 3)2
4 x + 24 + 45 x + 90
25(x + 2 )(x − 3)
2
=
=
25 x + 150
25(x + 2 )(x − 3)
2
(
) (
)
4 x 2 − 6 x + 9 − 4 x 2 − x − 6 + 45(x + 2 )
=
Example 1.3-12: Evaluate the integral ∫
25(x + 2 )(x − 3)2
25(x + 6 )
25(x + 2 )(x − 3)
x+5
3
x + 4x 2 + 4x
2
=
=
4 x 2 − 24 x + 36 − 4 x 2
25(x + 2 )(x − 3)2
x+6
(x + 2)(x − 3)2
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
Hamilton Education Guides
45
Advanced Integration
1.3 Integration by Partial Fractions
)
(
Second - Factor the denominator x 3 + 4 x 2 + 4 x into x x 2 + 4 x + 4 = x(x + 2)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the
denominator is repeated, the integrand can be represented in the following way:
x+5
3
2
x + 4x + 4x
=
(
x+5
2
x x + 4x + 4
=
)
x+5
x(x + 2 )
=
2
A
B
C
+
+
x x + 2 ( x + 2 )2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x+5
x 3 + 4x 2 + 4x
(
A (x + 2 )2 + Bx (x + 2 ) + Cx
=
x(x + 2 )(x + 2 )2
) (
)
x + 5 = A x 2 + 4 x + 4 + B x 2 + 2 x + Cx = Ax 2 + 4 Ax + 4 A + Bx 2 + 2 Bx + Cx
x + 5 = ( A + B )x 2 + (4 A + 2 B + C )x + 4 A
A+ B = 0
therefore,
4A = 5
4 A + 2B + C = 1
5
4
5
4
which result in having A = , B = − , and C = −
3
2
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
A
x+5
B
5
C
1
5
1
3
1
∫ x 3 + 4 x 2 + 4 x dx = ∫ x dx + ∫ x + 2 dx + ∫ (x + 2)2 dx = 4 ∫ x dx − 4 ∫ x + 2 dx − 2 ∫ (x + 2)2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
5
4
1
5
1
3
1
5
5
3
1
∫ x dx − 4 ∫ x + 2 dx − 2 ∫ (x + 2)2 dx = 4 ln x − 4 ln x + 2 + 2 ⋅ x + 2 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
5
4
5(x + 2 )2 − 5 x (x + 2 ) − 6 x
5 1 5 1
3
3 1
−
+0 =
+ c , then y ′ = ⋅ − ⋅
4 x 4 x + 2 2(x + 2 )2
2 x+2
4 x ( x + 2 )2
5
4
Let y = ln x − ln x + 2 + ⋅
=
5 x 2 + 20 x + 20 − 5 x 2 − 10 x − 6 x
(
4 x 3 + 4x 2 + 4x
)
=
(
4 x + 20
4 x 3 + 4x 2 + 4x
Example 1.3-13: Evaluate the integral ∫
)
=
1
5
x + 2x 4 + x 3
4(x + 5)
(
4 x 3 + 4x 2 + 4x
)
=
x+5
3
x + 4x 2 + 4x
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 5 + 2 x 4 + x 3 into x 3 x 2 + 2 x + 1 = x 3 (x + 1)2 .
Third - Write the linear factors in partial fraction form. Since both factors in the denominator
are repeated, the integrand can be represented in the following way:
Hamilton Education Guides
46
Advanced Integration
1.3 Integration by Partial Fractions
1
5
4
x + 2x + x
3
=
1
(
=
)
x 3 x 2 + 2x +1
1
x (x + 1)
3
A B
C
D
E
+
+
+
+
2
3
x x
x + 1 (x + 1)2
x
=
2
Fourth - Solve for the constants A , B , C , D , and E by equating coefficients of the like powers.
1
x 5 + 2x 4 + x
=
3
Ax 2 (x + 1)2 + Bx (x + 1)2 + C (x + 1)2 + Dx 3 (x + 1) + Ex 3
x 3 (x + 1)2
)
(
) (
(
)
1 = Ax 2 x 2 + 2 x + 1 + Bx x 2 + 2 x + 1 + C x 2 + 2 x + 1 + Dx 3 (x + 1) + Ex 3
1 = Ax 4 + 2 Ax 3 + Ax 2 + Bx 3 + 2 Bx 2 + Bx + Cx 2 + 2Cx + C + Dx 4 + Dx 3 + Ex 3
1 = ( A + D )x 4 + (2 A + B + D + E )x 3 + ( A + 2 B + C )x 2 + (B + 2C )x + C therefore,
2A + B + D + E = 0
A+ D = 0
A + 2B + C = 0
B + 2C = 0
C =1
which result in having A = 3 , B = −2 , C = 1 , D = −3 , and E = −1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
1
1
1
1
1
∫ x 5 + 2 x 4 + x 3 dx = 3∫ x dx − 2∫ x 2 dx + ∫ x 3 dx − 3∫ x + 1 dx − ∫ (x + 1)2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
1
1
2
1
1
1
∫ x dx − 2∫ x 2 dx + ∫ x 3 dx − 3∫ x + 1 dx − ∫ (x + 1)2 dx = 3 ln x + x − 2 x 2 − 3 ln x + 1 + x + 1 + c
3
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
2
x
Let y = 3 ln x + −
=
+
1
2x
2
− 3 ln x + 1 +
1 2
1
1
1
1
+
− 3⋅
−
+0
+ c , then y ′ = 3 ⋅ −
2
3
x x
x + 1 (x + 1)2
x +1
x
3x 2 (x + 1)2 − 2 x(x + 1)2 + (x + 1)2 − 3 x 3 (x + 1) − x 3
x 3 (x + 1)2
− 3x 3 − x 3
(
)
x 3 x 2 + 2x + 1
=
=
3x 4 + 3x 2 + 6 x 3 − 2 x 3 − 2 x − 4 x 2 + x 2 + 1 + 2 x − 3x 4
(
)
x 3 x 2 + 2x + 1
1
5
x + 2x 4 + x 3
Example 1.3-14: Evaluate the integral ∫
1
4
x − 6x 3 + 9x 2
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 4 − 6 x 3 + 9 x 2 into x 2 x 2 − 6 x + 9 = x 2 (x − 3)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the
denominator is repeated, the integrand can be represented in the following way:
Hamilton Education Guides
47
Advanced Integration
1.3 Integration by Partial Fractions
1
4
3
x − 6x + 9x
2
=
1
(
x 2 x 2 − 6x + 9
=
)
1
x (x − 3)
2
=
2
A B
C
D
+
+
+
2
x x
x − 3 (x − 3)2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
x 4 − 6x 3 + 9x
=
2
Ax (x − 3)2 + B (x − 3)2 + Cx 2 (x − 3) + Dx 2
x 2 (x − 3)2
(
) (
)
1 = Ax (x − 3)2 + B (x − 3)2 + Cx 2 (x − 3) + Dx 2 = Ax x 2 + 9 − 6 x + B x 2 + 9 − 6 x + Cx 2 (x − 3) + Dx 2
1 = Ax 3 + 9 Ax − 6 Ax 2 + Bx 2 + 9 B − 6 Bx + Cx 3 − 3Cx 2 + Dx 2
1 = ( A + C )x 3 + (− 6 A + B − 3C + D )x 2 + (9 A − 6 B )x + 9 B therefore,
−6 A + B − 3C + D = 0
A+C = 0
which result in having A =
9 A − 6B = 0
9B = 1
6
1
6
1
, B = , C = − , and D =
81
9
81
9
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
A
dx
B
C
6
D
dx
1
6
dx
dx
1
dx
∫ x 4 − 6 x 3 + 9 x 2 = ∫ x dx + ∫ x 2 dx + ∫ x − 3 dx + ∫ (x − 3)2 dx = 81 ∫ x + 9 ∫ x 2 − 81 ∫ x − 3 + 9 ∫ (x − 3)2
Sixth - Integrate each integral individually using integration methods learned in previous sections.
6 dx 1
+
81 x 9
∫
dx
6
dx
1
6
dx
1
6
1
∫ x 2 − 81 ∫ x − 3 + 9 ∫ (x − 3)2 = 81 ln x − 9 x − 81 ln x − 3 − 9(x − 3) + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
=
6
1
1
1
6 1
6
1
6
1
+
+0
⋅ +
− ⋅
ln x −
− ln x − 3 −
+ c , then y ′ =
2
81 x − 3 9(x − 3)2
81 x 9 x
81
9 x 81
9(x − 3)
6 x(x − 3)2 + 9(x − 3)2 − 6 x 2 (x − 3) + 9 x 2
81x (x − 3)
2
2
=
(
6 x 3 − 36 x 2 + 54 x + 9 x 2 − 54 x + 81 − 6 x 3 + 18 x 2 + 9 x 2
(
81x 2 x 2 − 6 x + 9
Hamilton Education Guides
)
) (
)
6 x x 2 − 6 x + 9 + 9 x 2 − 6 x + 9 − 6 x 3 + 18 x 2 + 9 x 2
(
81x 2 x 2 − 6 x + 9
=
(
81
81 x 4 − 6 x 3 + 9 x 2
)
)
=
1
4
x − 6x 3 + 9x 2
48
Advanced Integration
1.3 Integration by Partial Fractions
CASE III - The Denominator Has Distinct Quadratic Factors
In this case the quadratic factors of the form ax 2 + bx + c appear only once in the denominator and
are irreducible. To solve this class of rational fractions we equate each proper rational fraction
with a single fraction of the form
Ax + B
2
ax + bx + c
Cx + D
,
2
cx + dx + e
,
Ex + F
2
ex + fx + g
, etc. The following
examples show the steps as to how this class of integrals are solved.
x2 − x + 3
Example 1.3-15: Evaluate the integral ∫
dx .
x3 + x
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 + x into x x 2 + 1 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
x2 − x + 3
3
x +x
=
x2 − x + 3
(
)
x x 2 +1
=
A Bx + C
+
x x 2 +1
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x2 − x + 3
3
x +x
(
=
(
)
A x 2 + 1 + (Bx + C ) x
(
)
x x 2 +1
)
x 2 − x + 3 = A x 2 +1 + (Bx + C ) x = Ax 2 + A + Bx 2 + Cx
x 2 − x + 3 = ( A + B )x 2 + Cx + A
A+ B =1
therefore,
C = −1
A=3
which result in having A = 3 , B = −2 , and C = −1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x2 − x + 3
A
1
Bx + C
−2 x − 1
1
2x
1
∫ x 3 + x dx = ∫ x dx + ∫ x 2 + 1 dx = 3∫ x dx + ∫ x 2 + 1 dx = 3∫ x dx − ∫ x 2 + 1 dx − ∫ x 2 + 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous
sections. To solve the second integral let u = x 2 + 1 .
1
2x
1
1
2 x du
1
1
1
1
∫ x dx − ∫ x 2 + 1 dx − ∫ x 2 + 1 dx = 3∫ x dx − ∫ u ⋅ 2 x − ∫ x 2 + 1 dx = 3∫ x dx − ∫ u du − ∫ x 2 + 1 dx
3
= 3 ln x − ln u − tan −1 x + c = 3 ln x − ln x 2 + 1 − tan −1 x + c
Hamilton Education Guides
49
Advanced Integration
1.3 Integration by Partial Fractions
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
1
x
Let y = 3 ln x − ln x 2 + 1 − tan −1 x + c , then y ′ = 3 ⋅ +
=
(
)
3 x 2 +1 − 2x 2 − x
(
)
x x 2 +1
=
3x 2 + 3 − 2 x 2 − x
x3 + x
x +1
⋅ 2x −
1
1+ x
2
+0 =
3
2x
1
−
−
x x 2 +1 x 2 +1
x2 − x + 3
=
Example 1.3-16: Evaluate the integral ∫
−1
2
x3 + x
1
3
x + 25 x
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 + 25 x into x x 2 + 25 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
1
3
x + 25 x
=
1
(
x x 2 + 25
=
)
A Bx + C
+
x x 2 + 25
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
3
x + 25 x
(
=
(
)
A x 2 + 25 + (Bx + C ) x
(
x x 2 + 25
)
)
1 = A x 2 + 25 + (Bx + C ) x = Ax 2 + 25 A + Bx 2 + Cx
1 = ( A + B )x 2 + Cx + 25 A therefore,
25 A = 1
C=0
which result in having A =
A+ B = 0
1
1
, B = − , and C = 0
25
25
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
A
1
Bx + C
1
1
x
∫ x 3 + 25x dx = ∫ x dx + ∫ x 2 + 25 dx = 25 ∫ x dx − 25 ∫ x 2 + 25 dx
Sixth - Integrate each integral individually using integration methods learned in previous
sections. To solve the second integral let u = x 2 + 25 .
1
25
=
1
1
x
1
1
1
x du
1
1
1
1
1
1
∫ x dx − 25 ∫ x 2 + 25 dx = 25 ∫ x dx − 25 ∫ u ⋅ 2 x = 25 ∫ x dx − 50 ∫ u du = 25 ln x − 50 ln u + c
1
1
ln x −
ln x 2 + 25 + c
25
50
Hamilton Education Guides
50
Advanced Integration
1.3 Integration by Partial Fractions
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
1
1
1 1 1
1
1
−
⋅ 2x + 0 =
⋅ − ⋅
ln x −
ln x 2 + 25 + c , then y ′ =
2
25 x
25 x 50 x + 25
25
50
2
2
2
2
(
25 x 2 + 25
(x + 25)− x = x + 25 − x = 25 = 1
x + 25 x
25(x + 25 x )
25 x (x + 25)
25(x + 25 x )
=
x
)
2
3
3
3
Example 1.3-17: Evaluate the integral ∫
1
4
x + 16 x 2
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 4 + 16x 2 into x 2 x 2 + 16 .
Third - Write the factors in partial fraction form. Since the factors in the denominator are in
quadratic form, the integrand can be represented in the following way:
1
4
x + 16 x
=
2
1
(
x 2 x 2 + 16
=
)
Ax + B
x
2
+
Cx + D
x 2 + 16
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
1
x 4 + 16 x
(
=
2
( Ax + B ) (x 2 + 16)+ x 2 (Cx + D )
(
x 2 x 2 + 16
)
)
1 = ( Ax + B ) x 2 + 16 + x 2 (Cx + D ) = Ax 3 + 16 Ax + Bx 2 + 16 B + Cx 3 + Dx 2
1 = ( A + C )x 3 + (B + D )x 2 + 16 Ax + 16 B therefore,
B+D =0
A+C = 0
which result in having A = 0 , B =
16 A = 0
16 B = 1
1
1
, C = 0 , and D = −
16
16
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
Ax + B
B
Cx + D
1
D
1
1
1
∫ x 4 + 16 x 2 dx = ∫ x 2 dx + ∫ x 2 + 16 dx = ∫ x 2 dx + ∫ x 2 + 16 dx = 16 ∫ x 2 dx − 16 ∫ x 2 + 16 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
16
=
1
1
1
1
1
1
1
1
1
1
1
∫ x 2 dx − 16 ∫ x 2 + 16 dx = 16 ∫ x 2 dx − 16 ∫ 16(x 2 + 1 ) dx = 16 ∫ x 2 dx − 256 ∫ (x 2 + 1 ) dx
16
16
1
1
x
x
1
1
1
−
tan −1 + c
⋅− −
⋅ 4 tan −1 + c = −
4
16 x 256
16 x 64
4
Hamilton Education Guides
51
Advanced Integration
1.3 Integration by Partial Fractions
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = −
1
1
1
1
16
1
1
x
1
1
− ⋅
⋅ +0 =
−
⋅
−
tan −1 + c , then y ′ =
2
2
2
64 1 + x 4
256 16 + x 2
4
16 x 64
16 x
16 x
16
=
1
16 x 2
−
(
1
16 16 + x 2
=
)
(16 + x )− x = 16 + x − x =
16
1
=
x + 16 x
16 x (16 + x )
16 x (16 + x )
16 x (16 + x )
2
2
2
2
2
2
2
Example 1.3-18: Evaluate the integral ∫
1
3
x −8
2
2
4
2
2
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 − 8 into (x − 2) x 2 + 2 x + 4 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
1
3
x −8
=
1
(x − 2) (x 2 + 2 x + 4)
=
A
Bx + C
+
2
x − 2 x + 2x + 4
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
3
x −8
(
=
(
)
A x 2 + 2 x + 4 + (Bx + C )(x − 2 )
(x − 2) (x 2 + 2 x + 4)
)
1 = A x 2 + 2 x + 4 + (Bx + C )(x − 2 ) = Ax 2 + 2 Ax + 4 A + Bx 2 − 2 Bx + Cx − 2C
1 = ( A + B )x 2 + (2 A − 2 B + C )x + (4 A − 2C ) therefore,
A+ B = 0
4 A − 2C = 1
2 A − 2B + C = 0
which result in having A =
1
1
1
, B = − , and C = −
12
12
3
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
∫ x3 − 8
dx =
∫
A
dx +
x−2
Bx + C
∫ x 2 + 2x + 4
dx =
1
12
∫
1
dx +
x−2
1 x−1
− 12
3
1
1
1
x+4
∫ x 2 + 2 x + 4 dx = 12 ∫ x − 2 dx − 12 ∫ x 2 + 2 x + 4 dx
Sixth - Integrate each integral individually using integration methods learned in previous
sections. To solve the second integral let u = x 2 + 2 x + 4 , then
x + 4 can be rewritten as x + 4 = (x + 1) + 3 =
Hamilton Education Guides
du
du
. Also,
= 2 x + 2 and dx =
2x + 2
dx
1
(2 x + 2) + 3 . Therefore,
2
52
Advanced Integration
1
12
1.3 Integration by Partial Fractions
1
12
(x + 1) + 3 dx = 1
1
1
dx −
12
x−2
∫ x 2 + 2x + 4
12
3
1
1
1
2
(2 x + 2) + 3
∫
1
1
dx −
12
x−2
=
1
12
∫ x − 2 dx − 24 ∫ x 2 + 2 x + 4 dx − 12 ∫ x 2 + 2 x + 4 dx = 12 ∫ x − 2 dx − 24 ∫ u ⋅ 2 x + 2 − 12 ∫ (x + 1)2 + 3 dx
=
1
12
∫ x − 2 dx − 24 ∫ u du − 4 ∫ (x + 1)2 + 3 dx = 12 ln x − 2 − 24 ln u − 4 3 tan
=
1
1
3
3
x +1
x +1
1
1
ln x − 2 −
ln x 2 + 2 x + 4 −
tan −1
+c
ln x 2 + 2 x + 4 −
+c =
ln x − 2 −
tan −1
24
4⋅3
12
12
24
12
3
3
x+4
∫ x 2 + 2x + 4
1
1
1
1
dx =
2x + 2
1
∫
1
1
1
1
1
1
∫
1
1
dx −
x−2
12
2x + 2
1
∫ x 2 + 2 x + 4 dx
du
−1 x + 1
3
3
1
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
y′ =
1
1
3
x +1
ln x − 2 −
ln x 2 + 2 x + 4 −
tan −1
+ c , then
12
24
12
3
1
1
1
1
3
⋅
− ⋅
⋅ (2 x + 2 ) −
⋅
12 x − 2 24 x 2 + 2 x + 4
12
−
1
3
3
3
−
=
⋅
⋅
2
12(x − 2 )
3
12 3 + (x + 1)
=
1
+
12(x − 2 )
=
(
−x − 4
12 x 2 + 2 x + 4
4+8
(
12(x − 2 ) x + 2 x + 4
2
)
=
)
=
(
1
1 +  x +1 
 3
x +1
12 x 2 + 2 x + 4
)
−
⋅
2
(1⋅ 3 )− 0 ⋅ (x + 1) + 0 = 1 − x + 1
12(x − 2 )
( 3 )2
12(x 2 + 2 x + 4 )
3
(
12 x 2 + 2 x + 4
=
)
1
+
12(x − 2 )
(
−x −1− 3
12 x 2 + 2 x + 4
(x + 2 x + 4)+ (− x − 4)(x − 2) = x + 2 x + 4 − x + 2 x − 4 x + 8
12(x − 2 ) (x + 2 x + 4 )
12(x − 2 ) (x + 2 x + 4 )
2
2
2
2
(
12
12(x − 2 ) x + 2 x + 4
2
Example 1.3-19: Evaluate the integral ∫
2
)
1
3
)
x +8
=
1
3
2
2
x + 2x + 4x − 2x − 4x − 8
=
1
3
x −8
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 + 8 into (x + 2) x 2 − 2 x + 4 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
1
3
x +8
=
1
(x + 2) (x 2 − 2 x + 4)
=
A
Bx + C
+
2
x + 2 x − 2x + 4
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
3
x +8
Hamilton Education Guides
=
(
)
A x 2 − 2 x + 4 + (Bx + C )(x + 2 )
(x + 2) (x 2 − 2 x + 4)
53
Advanced Integration
1.3 Integration by Partial Fractions
)
(
1 = A x 2 − 2 x + 4 + (Bx + C )(x + 2 ) = Ax 2 − 2 Ax + 4 A + Bx 2 + 2 Bx + Cx + 2C
1 = ( A + B )x 2 + (− 2 A + 2 B + C )x + (4 A + 2C ) therefore,
A+ B = 0
−2 A + 2 B + C = 0
4 A + 2C = 1
1
1
1
, B = − , and C =
12
12
3
which result in having A =
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
∫ x3 + 8
dx =
∫
A
dx +
x+2
Bx + C
∫ x 2 − 2x + 4
dx =
1
12
1
dx +
x+2
∫
1 x+ 1
− 12
3
1
1
1
x−4
∫ x 2 − 2 x + 4 dx = 12 ∫ x + 2 dx − 12 ∫ x 2 − 2 x + 4 dx
Sixth - Integrate each integral individually using integration methods learned in previous
sections. To solve the second integral let u = x 2 − 2 x + 4 , then
x − 4 can be rewritten as x − 4 = (x − 1) − 3 =
1
12
x−4
du
du
. Also,
= 2 x − 2 and dx =
dx
2x − 2
1
(2 x − 2) − 3 . Therefore,
2
(x − 1) − 3 dx = 1
1
2
(2 x − 2) − 3
∫
1
1
dx −
12
x+2
=
1
12
∫ x + 2 dx − 24 ∫ x 2 − 2 x + 4 dx + 12 ∫ x 2 − 2 x + 4 dx = 12 ∫ x + 2 dx − 24 ∫ u ⋅ 2 x + 2 + 12 ∫ (x − 1)2 + 3 dx
=
1
12
∫ x + 2 dx − 24 ∫ u du + 4 ∫ (x − 1)2 + 3 dx = 12 ln x + 2 − 24 ln u + 4 3 tan
=
1
1
3
1
1
3
x −1
x −1
ln x + 2 −
ln x 2 − 2 x + 4 +
tan −1
ln x + 2 −
ln x 2 − 2 x + 4 +
tan −1
+c =
+c
12
24
4⋅3
12
24
12
3
3
∫ x 2 − 2x + 4
1
1
1
1
dx =
1
12
2x − 2
1
∫
1
1
dx −
x+2
12
∫ x 2 − 2x + 4
12 ∫
1
1
dx −
x+2
12
3
1
1
2x − 2
1
1
1
1
1
1
1
∫ x 2 − 2 x + 4 dx
du
−1 x − 1
3
3
1
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
y′ =
1
1
3
x −1
ln x + 2 −
ln x 2 − 2 x + 4 +
tan −1
+ c , then
12
24
12
3
1
1
1
1
3
⋅
−
⋅
⋅ (2 x − 2 ) +
⋅
2
12 x + 2 24 x − 2 x + 4
12
+
1
3
3
3
−
=
⋅
⋅
2
12(x + 2 )
12 3 + (x − 1)
3
=
1
+
12(x + 2 )
(
−x + 4
12 x 2 − 2 x + 4
Hamilton Education Guides
)
=
(
1
1 +  x −1 
 3
x −1
12 x 2 − 2 x + 4
)
+
2
(
⋅
(1⋅ 3 )− 0 ⋅ (x − 1) + 0 = 1 − x − 1
12(x + 2 )
( 3 )2
12(x 2 − 2 x + 4 )
3
12 x 2 − 2 x + 4
)
=
1
+
12(x + 2 )
(
−x +1+ 3
12 x 2 − 2 x + 4
(x − 2 x + 4)+ (− x + 4)(x + 2) = x − 2 x + 4 − x − 2 x + 4 x + 8
12(x + 2 ) (x − 2 x + 4 )
12(x + 2 ) (x − 2 x + 4 )
2
2
2
)
2
2
54
Advanced Integration
=
1.3 Integration by Partial Fractions
4+8
(
12(x + 2 ) x 2 − 2 x + 4
=
)
(
12
12(x + 2 ) x 2 − 2 x + 4
Example 1.3-20: Evaluate the integral ∫
1
=
)
x2
16 − x 4
3
2
2
x − 2x + 4x + 2x − 4x + 8
1
=
3
x +8
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)(
)
(
)
Second - Factor the denominator 16 − x 4 into 4 − x 2 4 + x 2 = (2 − x )(2 + x ) 4 + x 2 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
x2
16 − x
=
4
x2
A
B
Cx + D
+
+
2 − x 2 + x 4 + x2
=
(2 − x )(2 + x ) (4 + x 2 )
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
x2
16 − x
(
)
(
)
A(2 + x ) 4 − x 2 + B(2 − x ) 4 − x 2 + (2 − x )(2 + x )(Cx + D )
=
4
(
)
(2 − x )(2 + x ) (4 + x 2 )
(
)
x 2 = A(2 + x ) 4 + x 2 + B(2 − x ) 4 + x 2 + (2 − x )(2 + x )(Cx + D )
x 2 = 8 A + 2 Ax 2 + 4 Ax + Ax 3 + 8 B + 2 Bx 2 − 4 Bx − Bx 3 + 4Cx + 4 D − Cx 3 − Dx 2
x 2 = ( A − B − C )x 3 + (2 A + 2 B − D )x 2 + (4 A − 4 B + 4C )x + (8 A + 8 B + 4 D )
2 A + 2B − D = 1
A− B −C = 0
1
8
therefore,
4 A − 4 B + 4C = 0
1
8
which result in having A = , B = , C = 0 , and D = −
8 A + 8B + 4 D = 0
1
2
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x2
A
B
1
Cx + D
1
1
1
1
1
∫ 16 − x 4 dx = ∫ 2 − x dx + ∫ 2 + x dx + ∫ 4 + x 2 dx = 8 ∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 4 + x 2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
1
1
1
1
1
1
1
1
1
1
8
∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 4 + x 2 dx = 8 ∫ 2 − x dx + 8 ∫ 2 + x dx − 2 ∫ 2 2 + x 2 dx
=
1
1
1
1
1
x
1 1
x
ln 2 − x + ln 2 + x − ⋅ tan −1 + c = ln 2 − x + ln 2 + x − tan −1 + c
8
8
8
8
4
2
2 2
2
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Hamilton Education Guides
55
Advanced Integration
1.3 Integration by Partial Fractions
1
8
1
8
1
4
1 1
1 1
1
1
1
⋅
+ ⋅
− ⋅
⋅ +0
2
8 2 − x 8 2 + x 4 1+ x 2
x
2
Let y = ln 2 − x + ln 2 + x − tan −1 + c , then y ′ =
=
1
1
−
+
8(2 − x ) 8(2 + x )
4
(2 + x ) (4 + x 2 )+ (2 − x ) (4 + x 2 )− 4(2 − x )(2 + x ) 8 + 2 x 2 + 4 x + x 3 + 8
=
8(2 − x )(2 + x ) (4 + x 2 )
8(4 − x 2 )(4 + x 2 )
)
(2 x + 2 x + 4 x )+ (8 + 8 − 16) = 8x = x
2 x − 4 x − x − 16 + 4 x
=
+
16 − x
8(16 − x )
8(4 − x )(4 + x )
8(16 − x )
2
(
=
4
8 4 + x2
3
2
2
2
2
2
4
2
Example 1.3-21: Evaluate the integral ∫
4
5
4
2
2
x −1
4
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)(
)
(
)
Second - Factor the denominator x 4 − 1 into x 2 − 1 x 2 + 1 = (x − 1)(x + 1) x 2 + 1 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
5
4
x −1
=
5
(x − 1)(x + 1) (x 2 + 1)
=
A
B
Cx + D
+
+
x −1 x +1 x 2 +1
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
5
=
4
x −1
(
)
)
(
A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )
(
)
(x − 1)(x + 1) (x 2 + 1)
(
)
5 = A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )
5 = Ax 3 + Ax 2 + Ax + A + Bx 3 − Bx 2 + Bx − B + Cx 3 + Dx 2 − Cx − D
5 = ( A + B + C )x 3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D )
A+ B +C = 0
5
4
A− B + D = 0
A+ B −C = 0
5
4
5
2
which result in having A = , B = − , C = 0 , and D = −
therefore,
A− B − D = 5
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
5
A
B
Cx + D
5
1
5
1
5
1
∫ x 4 − 1 dx = ∫ x − 1 dx + ∫ x + 1 dx + ∫ x 2 + 1 dx = 4 ∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x 2 + 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
Hamilton Education Guides
56
Advanced Integration
5
4
1
5
1.3 Integration by Partial Fractions
1
5
5
1
5
5
∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x 2 + 1 dx = 4 ln x − 1 − 4 ln x + 1 − 2 tan x + c
−1
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
5
4
5
4
5
2
5 1
5 1
5
1
⋅
− ⋅
− ⋅
⋅1 + 0
2
4 x −1 4 x +1 2 x +1
Let y = ln x − 1 − ln x + 1 − tan −1 x + c , then y ′ =
=
5
5
−
−
4(x − 1) 4(x + 1)
5
(
(
2
)(
4 x −1 4 + x
2
)
2 x 2 +1
+ 5 x 2 − 5 x + 5 − 10 x 2 + 10
)
=
(
(
)
(
)
5(x + 1) x 2 + 1 − 5(x − 1) x 2 + 1 − 10(x − 1)(x + 1)
=
(
)
4(x − 1)(x + 1) x 2 + 1
20
4
)
4 x −1
=
=
5x 3 + 5x 2 + 5x + 5 − 5x 3
(
)(
4 x 2 −1 4 + x 2
)
5
4
x −1
Example 1.3-22: Evaluate the integral ∫
1
3
x − 64
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x 3 − 64 into (x − 4) x 2 + 4 x + 16 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator
is in quadratic form, the integrand can be represented in the following way:
1
3
x − 64
=
1
(x − 4) (x 2 + 4 x + 16)
=
A
Bx + C
+
2
x − 4 x + 4 x + 16
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
3
x − 64
=
(
)
(
)
A x 2 + 4 x + 16 + (Bx + C )(x − 4 )
(x − 4) (x 2 + 4 x + 16)
1 = A x 2 + 4 x + 16 + (Bx + C )(x − 4 ) = Ax 2 + 4 Ax + 16 A + Bx 2 − 4 Bx + Cx − 4C
1 = ( A + B )x 2 + (4 A − 4 B + C )x + (16 A − 4C ) therefore,
4 A − 4B + C = 0
A+ B = 0
which result in having A =
16 A − 4C = 1
1
1
1
, B = − , and C = −
48
6
48
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
∫ x 3 − 64
dx =
∫
A
dx +
x−4
Bx + C
∫ x 2 + 4 x + 16
dx =
1
48
∫
1
dx +
x−4
1 x− 1
− 48
6
1
1
1
x +8
∫ x 2 + 4 x + 16 dx = 48 ∫ x − 4 dx − 48 ∫ x 2 + 4 x + 16 dx
Sixth - Integrate each integral individually using integration methods learned in previous
Hamilton Education Guides
57
Advanced Integration
1.3 Integration by Partial Fractions
sections. To solve the second integral let u = x 2 + 4 x + 16 , then
Also, x + 8 can be rewritten as x + 8 = (x + 2) + 6 =
1
48
1
dx =
2
48
x + 4 x + 16
du
du
.
= 2 x + 4 and dx =
2x + 4
dx
1
(2 x + 4) + 6 . Therefore,
2
(x + 2) + 6 dx = 1
1
1
dx −
48
x−4
∫ x 2 + 4 x + 16
48
6
1
1
1
2
(2 x + 4) + 6
∫
1
1
dx −
48
x−4
=
1
48
∫ x − 4 dx − 96 ∫ x 2 + 4 x + 16 dx − 48 ∫ x 2 + 4 x + 16 dx = 48 ∫ x − 4 dx − 96 ∫ u ⋅ 2 x + 4 − 48 ∫ (x + 2)2 + 12 dx
=
1
48
∫ x − 4 dx − 96 ∫ u du − 8 ∫ (x + 2)2 + 12 dx = 48 ln x − 4 − 96 ln u − 8 12 tan
=
12
1
1
12
x+2
x+2
1
1
ln x − 4 −
ln x 2 + 4 x + 16 −
tan −1
+c
ln x − 4 − ln x 2 + 4 x + 16 −
tan −1
+c =
96
48
96
96
8 ⋅12
48
12
12
∫
1
1
1
1
x +8
2x + 4
1
∫
1
1
1
1
1
1
∫
1
1
dx −
x−4
48
2x + 4
1
∫ x 2 + 4 x + 16 dx
6
du
−1 x + 2
12
1
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
y′ =
1
1
12
x+2
ln x − 4 − ln x 2 + 4 x + 16 −
tan −1
+ c , then
48
96
96
12
12
1
1
1
1
⋅ (2 x + 4 ) −
⋅
⋅
− ⋅
96
48 x − 4 96 x 2 + 4 x + 16
−
12
12
12
⋅
⋅
2
96 12 + (x + 2 )
12
=
1
+
48(x − 4 )
=
(
=
−x − 8
48 x 2 + 4 x + 16
16 + 32
(
48(x − 4 ) x 2 + 4 x + 16
Hamilton Education Guides
)
)
=
1
−
48(x − 4 )
=
(
1
1 +  x + 2 
 12 
x+2
48 x 2 + 4 x + 16
)
⋅
2
−
(1⋅ 12 )− 0 ⋅ (x + 2) + 0 = 1 − x + 2
48(x − 4 )
( 12 )2
48(x 2 + 4 x + 16 )
6
(
48 x 2 + 4 x + 16
=
)
1
+
48(x − 4 )
(
−x − 2 − 6
48 x 2 + 4 x + 16
(x + 4 x + 16)+ (− x − 8)(x − 4) = x + 4 x + 16 − x + 4 x − 8x + 32
48(x − 4 ) (x + 4 x + 16 )
48(x − 4 ) (x + 4 x + 16 )
2
2
2
2
(
48
48(x − 4 ) x 2 + 4 x + 16
)
2
)
=
1
3
2
2
x + 4 x + 16 x − 4 x − 16 x − 64
=
1
3
x − 64
58
Advanced Integration
1.3 Integration by Partial Fractions
CASE IV - The Denominator Has Repeated Quadratic Factors
In this case each irreducible quadratic factor of the form ax 2 + bx + c appears n times in the
denominator. To solve this class of rational fractions we equate each proper rational fraction,
that appears n times in the denominator, with a sum of n partial fractions of the form
M 1 x + N1
2
ax + bx + c
+
M 2x + N2
( ax + bx + c)
2
2
+ ... +
M n x + Nn
( ax + bx + c)
2
n
. The following examples show the steps as to how
this class of integrals are solved.
x2
Example 1.3-23: Evaluate the integral ∫
x 4 + 2x 2 +1
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
2
Second - Factor the denominator x 4 + 2 x 2 + 1 into x 2 + 1 .
Third - Write the factors in partial fraction form. Since the quadratic form in the denominator
is repeated, the integrand can be represented in the following way:
x2
4
2
x + 2x + 1
=
x2
(x + 1)
2
2
=
Ax + B
+
2
x +1
Cx + D
(x + 1)
2
2
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
x2
x 4 + 2x 2 + 1
(
=
( Ax + B ) (x 2 + 1)+ Cx + D
)
(x + 1)
2
2
x 2 = ( Ax + B ) x 2 + 1 + Cx + D = Ax 3 + Ax + Bx 2 + B + Cx + D
x 2 = Ax 3 + Bx 2 + ( A + C )x + (B + D )
B =1
A=0
therefore,
A+C = 0
B+D =0
which result in having A = 0 , B = 1 , C = 0 , and D = −1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x2
Ax + B
0 +1
Cx + D
1
0 −1
1
∫ x 4 + 2 x 2 + 1 dx = ∫ x 2 + 1 dx + ∫ (x 2 + 1)2 dx = ∫ x 2 + 1 dx + ∫ (x 2 + 1)2 dx = ∫ x 2 + 1 dx − ∫ (x 2 + 1)2 dx
Sixth - Integrate each integral individually using integration methods learned in previous
sections. To solve the second integral let x = tan t , then
1
∫ x 2 +1
dx −
1
∫ (x 2 + 1)2
Hamilton Education Guides
dx = arc tan x −
sec 2 t
∫ (tan 2 t + 1)2
dx
= sec 2 t which implies dx = sec 2 t dt .
dt
dt = arc tan x −
sec 2 t
∫ (sec 2 t )2
dt = arc tan x −
sec 2 t
∫ sec 4 t dt
59
Advanced Integration
= arc tan x − ∫
1.3 Integration by Partial Fractions
1
2
sec t
∫
dt = arc tan x − cos 2 t dt = tan −1 x −
1
2
1
2
1
2
= tan −1 x − ( t + sin t cos t ) + c = tan −1 x − tan −1 x − ⋅
1
2
1
1

∫ ( 1 + cos 2t ) dt = tan x − 2  t + 2 sin 2t  + c
−1
x
1+ x 2
1
⋅
1+ x 2
+c =
x
1
+c
tan −1 x −
2
2 x2 + 1
(
)
Or, we could use the already derived integration formulas by using Tables 1.2-1 and 1.4-3.
Note – Since the objective of this section is to teach the process for solving integrals using the
Partial Fractions method, in the remaining example problems, we will use the already derived
integration formulas summarized primarily in Tables 1.2-1 and 1.4-3 in order to solve this
class of problems.
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
=
(
1
)
2 x 2 +1
−
(
)
+ c , then y ′ =
1
1
1 1⋅ x 2 + 1 − 2 x ⋅ x
1
+0 =
⋅
− ⋅
2
2 x 2 +1 2
x 2 +1
2 x 2 +1
− x 2 +1
x 2 +1+ x 2 −1
2x 2
2 x 2 +1
2 x 2 +1
1
2
Let y = tan −1 x −
(
x
(
)
2 x 2 +1
=
2
)
(
=
)
2
Example 1.3-24: Evaluate the integral ∫
(
=
2
)
2 x 2 +1
x 2 +1
x 4 + 8 x 2 + 16
(
x2
)
=
2
(x + 1)
2
(
)
−
x 2 − 2x 2 + 1
(
)
2 x 2 +1
2
x2
x 4 + 2x 2 + 1
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is an improper rational fraction use synthetic division (long division) to reduce the
rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
2
Second - Factor the denominator x 4 + 8 x 2 + 16 into x 2 + 4 .
Third - Write the factors in partial fraction form. Since the quadratic form in the denominator
is repeated, the integrand can be represented in the following way:
x 2 +1
4
2
x + 8 x + 16
x 2 +1
=
( x + 4)
2
2
=
Ax + B
2
x +4
+
Cx + D
(x + 4)
2
2
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
x 2 +1
x 4 + 8 x 2 + 16
(
)
=
( Ax + B ) (x 2 + 4)+ Cx + D
(x + 4)
2
2
x 2 + 1 = ( Ax + B ) x 2 + 4 + Cx + D = Ax 3 + 4 Ax + Bx 2 + 4 B + Cx + D
x 2 + 1 = Ax 3 + Bx 2 + (4 A + C )x + (4 B + D ) therefore,
A=0
B =1
4A + C = 0
4B + D = 1
which result in having A = 0 , B = 1 , C = 0 , and D = −3
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
Hamilton Education Guides
60
Advanced Integration
1.3 Integration by Partial Fractions
with their specific values.
x 2 +1
Ax + B
0 +1
Cx + D
0−3
3
1
∫ x 4 + 8x 2 + 16 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx − ∫ (x 2 + 4)2 dx
Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3.
1
1
3
∫ x 2 + 4 dx − ∫ (x 2 + 4)2 dx = 2 tan
3
x
− tan −1 −
2 16
2
−1 x
(
3x
2
8 x +4
)
+c =
x
5
tan −1 −
2
16
(
3x
2
8 x +4
)
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
(
x
5
tan −1 −
16
2
5
8 x2 + 4
)
(
3x
8 x2 + 4
)
+ c , then y ′ =
5
3 x 2 − 2x 2 + 4
=
− ⋅
2
8
8 x2 + 4
x2 + 4
(
(
)
)
(
)
5
1
1 3 1⋅ x 2 + 4 − 2 x ⋅ x
⋅
⋅ − ⋅
+0
2
16 x 2 + 1 2 8
2
x +4
2
()
3 − x2 + 4
− ⋅
2
8
x2 + 4
(
)
1
Example 1.3-25: Evaluate the integral ∫ 4
=
x + 10 x 2 + 25
(
)
5 x 2 + 20 + 3 x 2 − 12
(
8 x2 + 4
)
2
=
x 2 +1
=
2
(x + 4)
2
x 2 +1
x 4 + 8 x 2 + 16
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is a rational fraction use synthetic division (long division) to reduce the rational
fraction to the sum of a polynomial and a proper rational fraction.
(
)
2
Second - Factor the denominator x 4 + 10 x 2 + 25 into x 2 + 5 .
Third - Write the factors in partial fraction form. Since the quadratic form in the denominator
is repeated, the integrand can be represented in the following way:
1
4
=
2
x + 10 x + 25
1
( x + 5)
2
2
Ax + B
=
+
2
x +5
Cx + D
(x + 5)
2
2
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
1
x 4 + 10 x 2 + 25
(
)
=
( Ax + B ) (x 2 + 5)+ Cx + D
(x + 5)
2
2
1 = ( Ax + B ) x 2 + 5 + Cx + D = Ax 3 + 5 Ax + Bx 2 + 5 B + Cx + D
1 = Ax 3 + Bx 2 + (5 A + C )x + (5B + D ) therefore,
A=0
B=0
5A + C = 0
5B + D = 1
which result in having A = 0 , B = 0 , C = 0 , and D = 1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
Hamilton Education Guides
61
Advanced Integration
1.3 Integration by Partial Fractions
1
Ax + B
0+0
Cx + D
0 +1
1
∫ x4 + 10 x2 + 25 dx = ∫ x 2 + 5 dx + ∫ (x 2 + 5)2 dx = ∫ x 2 + 5 dx + ∫ (x 2 + 5)2 dx = ∫ (x 2 + 5)2 dx
Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3.
1
1
∫ (x 2 + 5)2 dx = 10 5 tan
x
−1
5
+
(
x
10 x 2 + 5
)
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
(
1
10 5
5
50 x 2 + 5
)
tan −1
+
x
5
+
x
(
10 x 2 + 5
+ c , then y ′ =
)
1 x 2 − 2x 2 + 5
1
=
⋅
2
10
x2 + 5
10 x 2 + 5
(
)
(
Example 1.3-26: Evaluate the integral ∫
)
+
1
1
⋅
2
10 5  x 
  +1
 5
− x2 + 5
(
10 x 2 + 5
x3
x 4 + 4x 2 + 4
⋅
1
+
5
x2 + 5− x2 + 5
=
2
)
(
10 x 2 + 5
(
)
1⋅ x 2 + 5 − 2 x ⋅ x
)
2
(
10 x 2 + 5
=
)
2
1
(x + 5)
2
2
+0
=
1
4
x + 10 x 2 + 25
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is a rational fraction use synthetic division (long division) to reduce the rational
fraction to the sum of a polynomial and a proper rational fraction.
(
)
2
Second - Factor the denominator x 4 + 4 x 2 + 4 into x 2 + 2 .
Third - Write the factors in partial fraction form. Since the quadratic form in the denominator
is repeated, the integrand can be represented in the following way:
x3
4
2
x + 4x + 4
=
x3
( x + 2)
2
2
=
Ax + B
2
x +2
+
Cx + D
(x + 2)
2
2
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
x3
x 4 + 4x 2 + 4
(
)
=
( Ax + B ) (x 2 + 2)+ Cx + D
(x + 2)
2
2
x 3 = ( Ax + B ) x 2 + 2 + Cx + D = Ax 3 + 2 Ax + Bx 2 + 2 B + Cx + D
x 3 = Ax 3 + Bx 2 + (2 A + C )x + (2 B + D ) therefore,
A =1
2A + C = 0
B=0
2B + D = 0
which result in having A = 1 , B = 0 , C = −2 , and D = 0
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
x3
Ax + B
Cx + D
x+0
−2 x + 0
x
2x
∫ x 4 + 4 x 2 + 4 dx = ∫ x 2 + 2 dx + ∫ (x 2 + 2)2 dx = ∫ x 2 + 2 dx + ∫ (x 2 + 2)2 dx = ∫ x 2 + 2 dx − ∫ (x 2 + 2)2 dx
Hamilton Education Guides
62
Advanced Integration
1.3 Integration by Partial Fractions
Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3.
2x
x
x 1
x
1
1
1
2
1
1
1
∫ x 2 + 2 dx − ∫ (x 2 + 2)2 dx = ∫ u ⋅ 2 x du − 2∫ u 2 ⋅ 2 x du = 2 ∫ u du − 2 ∫ u 2 du = 2 ∫ u du − ∫ u −2 du
=
1
1
1
1
1
1
1
ln u −
+c
u − 2+1 + c = ln u + u −1 + c = ln u + + c = ln x 2 + 2 +
2
2
2
− 2 +1
2
2
u
x +2
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
1
2
Let y = ln x 2 + 2 +
=
x 3 + 2x − 2x
(x + 2)
2
2
=
1
2
x +2
x3
+ c , then y ′ =
=
2
(x + 2)
2
1 2x
⋅
−
2 x2 + 2
2x
(x + 2)
2
2
+0 =
x
2
x +2
−
2x
=
2
(x + 2)
2
(
)
x x 2 + 2 − 2x
(x + 2)
2
2
x3
x 4 + 4x 2 + 4
Example 1.3-27: Evaluate the integral ∫
2x 2 + x + 7
x 4 + 8 x 2 + 16
dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the
integrand is a rational fraction use synthetic division (long division) to reduce the rational
fraction to the sum of a polynomial and a proper rational fraction.
(
)
2
Second - Factor the denominator x 4 + 8 x 2 + 16 into x 2 + 4 .
Third - Write the factors in partial fraction form. Since the quadratic form in the denominator
is repeated, the integrand can be represented in the following way:
2x 2 + x + 7
4
=
2
x + 8 x + 16
2x 2 + x + 7
( x + 4)
2
2
=
Ax + B
2
x +4
+
Cx + D
(x + 4)
2
2
Fourth - Solve for the constants A , B , C , and D by equating coefficients of the like powers.
2x 2 + x + 7
x 4 + 8 x 2 + 16
(
=
)
( Ax + B ) (x 2 + 4)+ Cx + D
(x + 4)
2
2
2 x 2 + x + 7 = ( Ax + B ) x 2 + 4 + Cx + D = Ax 3 + 4 Ax + Bx 2 + 4 B + Cx + D
2 x 2 + x + 7 = Ax 3 + Bx 2 + (4 A + C )x + (4 B + D ) therefore,
B=2
A=0
4A + C = 1
4B + D = 7
which result in having A = 0 , B = 2 , C = 1 , and D = −1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
2x 2 + x + 7
Ax + B
Cx + D
0+2
x −1
2
x −1
∫ x 4 + 8x 2 + 16 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx
= ∫
2
2
x +4
dx +
x
1
∫ (x 2 + 4)2 dx − ∫ (x 2 + 4)2 dx
Hamilton Education Guides
63
Advanced Integration
1.3 Integration by Partial Fractions
Sixth - Integrate each integral individually by using Tables 1.2-1 and 1.4-3.
2
1
x
1
1
1
x
x
x
x
∫ x 2 + 4 dx + ∫ (x 2 + 4)2 dx − ∫ (x 2 + 4)2 dx = 2 ⋅ 2 tan −1 2 + ∫ u 2 ⋅ 2 x du − 16 tan −1 2 − 8(x2 + 4) + c
1
2
x
1
1
1
u − 2+1 − tan −1 −
2
2 − 2 +1
16
x
2
= 2 ⋅ tan −1 + ⋅
x
2
= tan −1 −
(
1
2
2 x +4
)
−
x
1
tan −1 −
16
2
(
x
2
8 x +4
x
8 x2 + 4
)
+c =
15
x
tan −1 +
16
2
)
(
+ c = tan −1
(
x 1
x
1
−
− tan −1 −
2
2 2u 16
−4 − x
2
8 x +4
)
+c =
x
(
8 x2 + 4
x
15
tan −1 −
16
2
)
(
+c
x+4
2
8 x +4
)
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
15
x
tan −1 −
16
2
15
x+4
(
2
8 x +4
)
+ c , then y ′ =
1 x 2 − 2 x 2 − 8x + 4
15
=
− ⋅
2
8
x2 + 4
8 x2 + 4
( )
( )
( )
8 (2 x 2 + x + 7 )
2x 2 + x + 7
2x + x + 7
=
=
=
2
8(x 2 + 4 )
(x + 4) x 4 + 8x + 16
8 x2 + 4
(
)
15
1
1 1 ⋅ x 2 + 4 − 2 x ⋅ (x + 4)
⋅
⋅ −
+0
2
16 x 2 + 1 2
2
8 x +4
2
()
+
x 2 + 8x − 4
(
8 x2 + 4
)
2
=
(
)
15 x 2 + 60 + x 2 + 8 x − 4
(
8 x2 + 4
)
2
=
16 x 2 + 8 x + 56
(
8 x2 + 4
)
2
2
2
2
Section 1.3 Practice Problems – Integration by Partial Fractions
Evaluate the following integrals:
a. ∫
d. ∫
g. ∫
dx
2
x + 5x + 6
=
b. ∫
dx =
e. ∫
x+5
x 3 + 2x 2 + x
1
3
x −1
dx =
Hamilton Education Guides
x 2 +1
3
x − 4x
dx =
1
x 3 − 2x 2 + x
1
h. ∫ 4
x −1
dx =
dx =
c. ∫
f. ∫
i. ∫
1
36 − x 2
x2 + 3
x 2 −1
dx =
1
3
dx =
x + 64
dx =
64
Advanced Integration
1.4
1.4 Integration of Hyperbolic Functions
Integration of Hyperbolic Functions
In the following examples we will solve problems using the formulas below:
Table 1.4-1: Integration Formulas for Hyperbolic Functions
1. ∫ sinh x dx = cosh x + c
2. ∫ cosh x dx = sinh x + c
3. ∫ tanh x dx = ln cosh x + c
4. ∫ coth x dx = ln sinh x + c
5. ∫ sec h x dx = sin −1 (tanh x ) + c
6. ∫ csc h x dx = ln tanh
7. ∫ tanh x sec h x dx = − sec h x + c
8. ∫ coth x csc h x dx = − csc h x + c
9. ∫ sinh 2 x dx =
11. ∫ tanh 2 x dx = x − tanh x + c
12. ∫ coth 2 x dx = x − coth x + c
10. ∫ cosh 2 x dx =
sinh 2 x x
+ +c
4
2
13. ∫ sec h 2 x dx = tanh x + c
x
+c
2
sinh 2 x x
− +c
4
2
14. ∫ csc h 2 x dx = − coth x + c
Additionally, the following formulas, similar to the trigonometric functions, hold for the
hyperbolic functions:
1. Unit Formulas
tanh h 2 x + sec h 2 x = 1
cosh 2 x − sinh 2 x = 1
coth h 2 x − csc h 2 x = 1
2. Addition Formulas
sinh (x ± y ) = sinh x cosh y ± cosh x sinh y
tanh (x ± y ) =
tanh x ± tanh y
1 ± tanh x tanh y
cosh (x ± y ) = cosh x cosh y ± sinh h x sinh y
coth (x ± y ) =
coth x coth y ± 1
coth y ± coth x
3. Half Angle Formulas
sinh
1
x=
2
cosh x − 1
2
cosh
1
x=
2
cosh x + 1
2
tanh
1
x=
2
cosh x − 1
cosh x + 1
tanh
sinh x
cosh x − 1
1
=
x =
cosh x + 1
sinh x
2
x 0
sinh
cosh x − 1
1
x=−
2
2
x 0
x 0
sinh
cosh x − 1
1
x=−
2
cosh x + 1
x 0
or,
4. Double Angle Formulas
sinh 2 x = 2 sinh x cosh x
tanh 2 x =
cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 x − 1 = 1+ 2 sinh 2 x
2 tanh x
1 + tanh 2 x
Hamilton Education Guides
65
Advanced Integration
1.4 Integration of Hyperbolic Functions
Also, the hyperbolic functions are defined by
)
(e − e ) = e − e
sinh x
=
tanh x =
cosh x
(e + e ) e + e
sinh x =
sec h x =
(
)
(e + e ) = e + e
cosh x
=
coth x =
sinh x
(e − e ) e − e
1 x
e − e −x
2
cosh x =
1
2
x
−x
x
−x
1
2
x
−x
x
−x
1
=
cosh x 1
2
1
(e + e )
x
−x
=
2
x
e +e
csc h x =
−x
(
1 x
e + e −x
2
1
2
x
−x
x
−x
1
2
x
−x
x
−x
1
=
sinh x 1
2
1
(e − e )
x
−x
=
2
x
e − e −x
Also note that the negative argument of the hyperbolic functions is equal to the following:
sinh (− x ) = − sinh x
tanh (− x ) = − tanh x
coth (− x ) = − coth x
csc h (− x ) = − csc h x
cosh (− x ) = cosh x
sec h (− x ) = sec h x
Finally, we need to know how to differentiate the hyperbolic functions (addressed in Chapter 3,
Section 3.4 of Calculus 1 book) in order to check the answer to the given integrals below. The
derivatives of hyperbolic functions are repeated here and are as follows:
Table 1.4-2: Differentiation Formulas for Hyperbolic Functions
d
du
coth u = − csc h 2 u ⋅
dx
dx
du
d
sec h u = − sec h u tanh u ⋅
dx
dx
du
d
csc h u = − csc h u coth u ⋅
dx
dx
du
d
sinh u = cosh u ⋅
dx
dx
d
du
cosh u = sinh u ⋅
dx
dx
d
du
tanh u = sec h 2 u ⋅
dx
dx
Let’s integrate some hyperbolic functions using the above integration formulas.
Example 1.4-1: Evaluate the following integrals:
1
6
b. ∫ sinh x dx =
c. ∫ cosh 7 x dx =
d. ∫ cosh x dx =
e. ∫ (sinh 4 x + cosh 2 x ) dx =
f. ∫ csc h 8 x dx =
g. ∫ csc h 2 5 x dx =
h. ∫ csc h 2
a. ∫ sinh 5 x dx =
1
5
(
)
j. ∫ x sec h 2 x 2 + 5 dx =
1
x dx =
4
i. ∫ x 2 sec h 2 x 3 dx =
k. ∫ x 2 csc h 2 x 3 dx =
l. ∫ 2 x csc h 2 x 2 dx =
Solutions:
a. Given ∫ sinh 5 x dx let u = 5 x , then
du
1
du
du d
. Therefore,
=
5 x = 5 which implies dx =
5
dx dx
1
1
∫ sinh 5x dx = ∫ sinh u ⋅ 5 = 5 ∫ sinh u du = 5 cosh u + c = 5 cosh 5 x + c
Hamilton Education Guides
66
Advanced Integration
1.4 Integration of Hyperbolic Functions
1
5
Check: Let y = cosh 5 x + c , then y ′ =
1 d
d
1
d
1
⋅ sinh 5 x +
c = ⋅ sinh 5 x ⋅
5 x + 0 = ⋅ sinh 5 x ⋅ 5
5 dx
dx
5
dx
5
5
⋅ sinh 5 x = sinh 5 x
5
x
du d x
x
1
= which implies dx = 6 du . Therefore,
=
sinh dx let u = , then
6
dx dx 6
6
6
=
b. Given ∫
x
x
∫ sinh 6 dx = ∫ sinh u ⋅ 6 du = 6∫ sinh u du = 6 cosh u + c = 6 cosh 6 + c
x
6
Check: Let y = 6 cosh + c , then y ′ = 6 ⋅
=
d
x d
x 1
x d x
cosh +
c = 6 ⋅ sinh ⋅
+ 0 = 6 ⋅ sinh ⋅
6 dx
dx
6 dx 6
6 6
x
6
x
= sinh
⋅ sinh
6
6
6
c. Given ∫ cosh 7 x dx let u = 7 x , then
du
du d
du
. Therefore,
=
7 x = 7 which implies dx =
dx dx
7
1
1
1
∫ cosh 7 x dx = ∫ cosh u ⋅ 7 = 7 ∫ cosh u du = 7 sinh u + c = 7 sinh 7 x + c
1
7
Check: Let y = sinh 7 x + c , then y ′ =
1 d
d
1
1
d
⋅ sinh 7 x +
c = ⋅ cosh 7 x ⋅
7 x + 0 = ⋅ cosh 7 x ⋅ 7
7 dx
dx
7
7
dx
7
⋅ cosh 7 x = cosh 7 x
7
1
x
x
du d x
= which implies dx = 5 du . Therefore,
cosh dx let u = , then
=
5
5
5
dx dx 5
=
d. Given ∫
x
x
∫ cosh 5 dx = ∫ cosh u ⋅ 5 du = 5∫ cosh u du = 5 sinh u + c = 5 sinh 5 + c
x
5
Check: Let y = 5 sinh + c , then y ′ = 5 ⋅
=
d
x d
x d x
x 1
sinh +
+ 0 = 5 ⋅ cosh ⋅
c = 5 ⋅ cosh ⋅
5 dx
dx
5 5
5 dx 5
5
x
x
⋅ cosh = cosh
5
5
5
e. Given ∫ (sinh 4 x + cosh 2 x ) dx = ∫ sinh 4 x dx + ∫ cosh 2 x dx let:
a. u = 4 x , then
du
du d
du
and
=
4x ;
= 4 ; du = 4dx ; dx =
4
dx dx
dx
b. v = 2 x , then
dv
dv
dv d
.
=
2x ;
= 2 ; dv = 2dx ; dx =
2
dx dx
dx
Therefore, ∫ sinh 4 x dx + ∫ cosh 2 x dx = ∫ sinh u ⋅
=
1
1
du
dv
sinh u du +
cosh v dv
+ cosh v ⋅
=
4
2
4
2
∫
∫
∫
1
1
1
1
1
1
cosh u + c1 + sinh v + c 2 = cosh 4 x + sinh 2 x + c1 + c 2 = cosh 4 x + sinh 2 x + c
4
2
4
2
4
2
Hamilton Education Guides
67
Advanced Integration
1.4 Integration of Hyperbolic Functions
1
4
1
2
Check: Let y = cosh 4 x + sinh 2 x + c then y ′ =
1 d
1 d
d
d
1
⋅ cosh 4 x + ⋅ sinh 2 x +
c = ⋅ sinh 4 x ⋅
4x
4 dx
2 dx
dx
dx
4
4
2
1
d
2 x + 0 = ⋅ sinh 4 x + ⋅ cosh 2 x = sinh 4 x + cosh 2 x
⋅ cosh 2 x ⋅
4
2
2
dx
du d
du
csc h 8 x dx let u = 8 x , then
. Therefore,
=
8 x = 8 which implies du = 8dx ; dx =
dx dx
8
+
f. Given ∫
du
1
1
u
1
1
8x
∫ csc h 8x dx = ∫ csc h u ⋅ 8 = 8 ∫ csc h u du = 8 ln tanh 2 + c = 8 ln tanh 2 + c = 8 ln tanh 4 x + c
d
1
1
1 d
d
⋅ (tanh 4 x ) + 0
⋅ ln tanh 4 x +
c = ⋅
8 tanh 4 x dx
8 dx
dx
1
8
Check: Let y = ln tanh 4 x + c , then y ′ =
=
(
2
2
2
1
1
1
1 4 sec h 4 x
1 sec h 4 x
1 1 − tanh 4 x
= ⋅
= ⋅
= ⋅
⋅
⋅ sec h 2 4 x ⋅ 4 + 0 = ⋅
8 tanh 4 x
8 tanh 4 x
2 tanh 4 x
tanh 4 x
2
2
1
= ⋅
2
=
)
2
cosh 2 4 x − sinh 2 4 x
cosh 2 4 x
sinh 4 x
cosh 4 x
1 − sinh 2 4 x
cosh 4 x
sinh 4 x
cosh 4 x
1
2
cosh 4 x
1
1
1
1
4x
= ⋅ cosh
= ⋅
=
=
2 cosh 2 4 x ⋅ sinh 4 x
2 cosh 4 x ⋅ sinh 4 x
sinh 2 ⋅ 4 x
2 sinh 4 x
cosh 4 x
1
= csc h 8 x
sinh 8 x
g. Given ∫ csc h 2 5 x dx let u = 5 x , then
du
du
du d
du
. Therefore,
= 5 ; du = 5dx ; dx =
=
5x ;
dx
5
dx dx
1
1
1
∫ csc h 5x dx = ∫ csc h u ⋅ 5 = 5 ∫ csc h u du = − 5 coth u + c = − 5 coth 5 x + c
2
2
1
5
2
(
)
1 d
(coth 5 x )⋅ d 5 x + d c = − 1 ⋅ − csc h 2 5 x ⋅ 5 + 0
dx
dx
5
5 dx
Check: Let y = − coth 5 x + c , then y ′ = − ⋅
5
⋅ csc h 2 5 x = csc h 2 5 x
5
1
1
du d x du 1
;
csc h 2 x dx let u = x , then
= ; 4du = dx ; dx = 4du . Therefore,
=
4
4
dx dx 4 dx 4
=
h. Given ∫
1
2 1
∫ csc h 4 x dx = ∫ csc h 2 u ⋅ 4du = 4∫ csc h 2 u du = − 4 coth u + c = − 4 coth 4 x + c
x
4
Check: Let y = −4 coth + c , then y ′ = − 4 ⋅
=
x 1
x d
d
x d x
c = − 4 ⋅ − csc h 2 ⋅
+ 0 = 4 csc h 2 ⋅
coth +
4 4
4 dx 2
dx
4 dx
4
x
x
1
= csc h 2 = csc h 2 x
csc h 2
4
4
4
4
i. Given ∫ x 2 sec h 2 x 3 dx let u = x 3 , then
Hamilton Education Guides
du
du d 3 du
. Therefore,
=
x ;
= 3x 2 ; du = 3 x 2 dx ; dx =
dx
dx dx
3x 2
68
Advanced Integration
1.4 Integration of Hyperbolic Functions
du
1
1
1
∫ x sec h x dx = ∫ x sec h u ⋅ 3x 2/ = 3 ∫ sec h 2 u du = 3 tanh u + c = 3 tanh x + c
2
2 3
2
2
1
3
Check: Let y = tanh x 3 + c , then y ′ =
=
3x 2
⋅ sec h 2 x 3 = x 2 sec h 2 x 3
3
(
3
1
d
1
1 d
d 3
c = ⋅ sec h 2 x 3 ⋅
⋅ tanh x 3 +
x + 0 = ⋅ sec h 2 x 3 ⋅ 3 x 2
3
dx
3
3 dx
dx
)
j. Given ∫ x sec h 2 x 2 + 5 dx let u = x 2 + 5 , then
(
)
du
du
du d 2
. Therefore,
= 2 x ; du = 2 x dx ; dx =
=
x +5 ;
dx
dx dx
2x
∫ x sec h (x + 5) dx = ∫ x sec h 2 u ⋅ 2 x = 2 ∫ sec h u du = 2 tanh u + c = 2 tanh ( x + 5)+ c
2
1
du
2
1
2
(
)
Check: Let y = tanh x 2 + 5 + c , then y ′ =
)
(
(
2
)
(
) (
)
1 d
d
1
d
⋅ tanh x 2 + 5 +
c = ⋅ sec h 2 x 2 + 5 ⋅
x2 + 5 + 0
2 dx
dx
2
dx
)
(
)
1
2x
⋅ sec h 2 x 2 + 5 ⋅ 2 x =
⋅ sec h 2 x 2 + 5 = x sec h 2 x 2 + 5
2
2
du
du d 3 du
x 2 csc h 2 x 3 dx let u = x 3 , then
. Therefore,
=
= 3x 2 ; du = 3 x 2 dx ; dx =
x ;
dx dx
dx
3x 2
=
k. Given ∫
(
1
1
2
du
1
1
1
∫ x csc h x dx = ∫ x csc h u ⋅ 3x 2 = 3 ∫ csc h u du = − 3 coth u + c = − 3 coth x + c
2
2 3
2
2
1
3
2
3
1
d 3
1 d
d
coth x 3 +
c = − ⋅ − csc h 2 x 3 ⋅
x +0
3
dx
dx
3 dx
Check: Let y = − coth x 3 + c , then y ′ = − ⋅
3x 2
1
⋅ csc h 2 x 3 = x 2 csc h 2 x 3
⋅ csc h 2 x 3 ⋅ 3 x 2 =
3
3
du
du d 2 du
2 x csc h 2 x 2 dx let u = x 2 , then
. Therefore,
= 2 x ; du = 2 x dx ; dx =
=
x ;
2x
dx
dx dx
=
l. Given ∫
du
∫ 2 x csc h x dx = ∫ 2 x csc h 2 u ⋅ 2 x = ∫ csc h 2 u du = − coth u + c = − coth x + c
2 2
2
Check: Let y = − coth x 2 + c , then y ′ = −
d
d
d 2
x + 0 = csc h 2 x 2 ⋅ 2 x
coth x 2 +
c = csc h 2 x 2 ⋅
dx
dx
dx
= 2 x csc h2 x 2
Example 1.4-2: Evaluate the following indefinite integrals:
a. ∫ sec h 2 5 x dx =
b. ∫ x 2 sinh x 3 dx =
c. ∫ x csc h x 2 dx =
d. ∫ sinh 5 (x + 1) cosh (x + 1) dx =
e. ∫ cosh 5 x sinh x dx =
f. ∫ cosh 5 5 x sinh 5 x dx =
h. ∫ x 2 cosh x 3 dx =
i. ∫
k. ∫ csc h 7 x coth 7 x dx =
l. ∫ tanh 2 10 x dx =
n. ∫ sinh 3 x dx =
o. ∫ sinh x dx =
g. ∫ cosh 4
x
x
sinh dx =
2
2
j. ∫ sec h 2 (5 x − 1) dx =
m. ∫ cosh 3
x
dx =
5
Hamilton Education Guides
e 2x
sec h e 2 x dx =
3
x
2
69
Advanced Integration
1.4 Integration of Hyperbolic Functions
Solutions:
du
du d
du
. Therefore,
=
5x ;
= 5 ; du = 5 dx ; dx =
dx
dx dx
5
a. Given ∫ sec h 2 5 x dx let u = 5 x , then
1
du
1
1
∫ sec h 5x dx = ∫ sec h 2 u ⋅ 5 = 5 ∫ sec h 2 u du = 5 tanh u + c = 5 tanh 5 x + c
2
1 d
d
1
5
⋅ tanh 5 x +
c = sec h 2 5 x ⋅ 5 + 0 = sec h 2 5 x = sec h 2 5 x
5 dx
dx
5
5
1
5
Check: Let y = tanh 5 x + c , then y ′ =
b. Given ∫ x 2 sinh x 3 dx let u = x 3 , then
Thus,
du
du
du d 3
.
=
x = 3x 2 which implies du = 3 x 2 dx ; dx =
dx dx
3x 2
1
1
1
∫ x sinh x dx = ∫ x sinh u ⋅ 3x 2 = 3 ∫ sinh u du = 3 cosh u + c = 3 cosh x + c
2
3
2
1
3
Check: Let y = cosh x 3 + c , then y ′ =
=
1
d 3
1
d
1 d
x + 0 = ⋅ sinh x3 ⋅ 3 x 2
c = ⋅ sinh x 3 ⋅
⋅ cosh x 3 +
dx
3
dx
3 dx
3
3x 2
⋅ sinh x3 = x 2 sinh x3
3
c. Given ∫ x csc h x 2 dx let u = x 2 , then
∫
3
x csc h x 2 dx =
∫
x csc h u ⋅
1
2
Check: Let y = ln tanh
du
2x
=
1
2
∫
du
du d 2 du
. Therefore,
= 2 x ; du = 2 x dx ; dx =
=
x ;
2x
dx dx
dx
csc h u du =
x2
1
1
u
+c
ln tanh
+ c = ln tanh
2
2
2
2
1
1
d
x2
1 d
x2
x2
d
tanh
⋅
+0
+ c , then y ′ = ⋅ ln tanh
+ c = ⋅
2 tanh x 2 dx
2
2
2 dx
2
dx
2
2 x2
2 x2
2
d x2
1
1
x
1 sec h 2 2 x
x 1 − tanh 2
2 x
h
⋅
⋅
+0 = ⋅
sec
= ⋅
= ⋅
= ⋅
⋅
2
2
2
2 tanh x
2 dx 2
2
2 tanh x
2
2
tanh x
2
2
1−
sinh 2 x
cosh
2
sinh x
2
x2
2
cosh
=
2
2
cosh 2 x − sinh 2 x
2
2
2
cosh 2 x
x
2
⋅
x2
2
sinh
cosh
=
x
sinh x 2
1
=
2
x2
2 x2
x cosh 2
⋅
x2
2
sinh
cosh
2
= x csc h x
2
x2
=
x
⋅
2
2
cosh
cosh 2
Hamilton Education Guides
2
x
x
⋅ sinh
2
2
x
=
2 cosh
2
2
x
x
⋅ sinh
2
2
=
2
2
x
2
sinh 2 ⋅ x2
2
d. Given ∫ sinh 5 (x + 1) cosh (x + 1) dx let u = sinh (x + 1) , then
du = cosh (x + 1) ⋅ dx ; dx =
2
x2
2
2
2
2 x2
du d
du
=
sinh (x + 1) ;
= cosh (x + 1) ;
dx dx
dx
du
. Therefore,
cosh (x + 1)
70
Advanced Integration
1.4 Integration of Hyperbolic Functions
du
1
1
∫ sinh (x + 1) cosh (x + 1) dx = ∫ u cosh (x + 1)⋅ cosh (x + 1) = ∫ u du = 6 u 6 + c = 6 sinh (x + 1) + c
5
5
6
1
⋅ 6 sinh 5 (x + 1) ⋅ cosh (x + 1) + 0 = sinh 5 (x + 1) cosh (x + 1)
6
1
6
Check: Let y = sinh 6 (x + 1) + c , then y ′ =
du
du d
du
. Therefore,
=
cosh x ;
= sinh x ; dx =
sinh x
dx dx
dx
e. Given ∫ cosh 5 x sinh x dx let u = cosh x , then
du
5
1
1
∫ cosh x sinh x dx = ∫ u sinh x ⋅ sinh x = ∫ u du = 6 u 6 + c = 6 cosh x + c
5
5
5
1
6
Check: Let y = cosh 6 x + c , then y ′ =
6
6
1
⋅ 6 cosh 5 x ⋅ sinh x + 0 = cosh 5 x sinh x = cosh 5 x sinh x
6
6
du
du
du d
. Thus,
=
cosh 5 x ;
= 5 sinh 5 x ; dx =
5 sinh 5 x
dx
dx dx
f. Given ∫ cosh 5 5 x sinh 5 x dx let u = cosh 5 x , then
du
1
1
1 1
∫ cosh 5x sin 5 5x dx = ∫ u sinh 5x ⋅ 5 sinh 5x = 5 ∫ u 5 du = 5 ⋅ 6 u 6 + c = 30 cosh 5 x + c
5
5
Check: Let y =
g. Given ∫ cosh 4
6 cosh 5 5 x ⋅ 5 sinh 5 x
1
30
+0 =
cosh 6 5 x + c , then y ′ =
cosh 5 5 x sinh 5 x = cosh 5 hx sinh 5 x
30
30
30
x
x
x
2du
x
du d
x du 1
sinh dx let u = cosh , then
. Therefore,
= sinh ; dx =
=
cosh ;
2
2
2
2 dx 2
2
dx dx
sin x
2
x
4 x
6
2 du
x
2
2
5 x
∫ cosh 2 sinh 2 dx = ∫ u sinh 2 ⋅ sinh x = 2∫ u du = 5 u + c = 5 cosh 2 + c
4
4
5
2
2
5
x
2
Check: Let y = cosh 5 + c , then y ′ =
h. Given ∫ x 2 cosh x 3 dx let u = x 3 , then
du
x
x
x
x 1
2
10
x
x
⋅ 5 cosh 4 ⋅ sinh ⋅ + 0 =
cosh 4 sin = cosh 4 sinh
2
2
5
2
2 2
2
2
10
du
du d 3 du
. Therefore,
=
= 3x 2 ; du = 3 x 2 ⋅ dx ; dx =
x ;
dx dx
dx
3x 2
1
1
1
∫ x cosh x dx = ∫ x cosh u ⋅ 3x 2 = 3 ∫ cosh u du = 3 ⋅ sinh u + c = 3 sinh x + c
2
3
2
1
3
Check: Let y = sinh x 3 + c , then y ′ =
i. Given ∫
∫
3
3
1
⋅ cosh x 3 ⋅ 3 x 2 + c = x 2 cosh x 3 + 0 = x 2 cosh x 3
3
3
du
e 2x
du d 2 x du
. Thus,
sec h e 2 x dx let u = e 2 x , then
= 2e 2 x ; du = 2e 2 x ⋅ dx ; dx =
=
e ;
3
dx dx
dx
2e 2 x
1 2x
du
e 2x
e sec h u ⋅
sec h e 2 x dx =
3
3
2e 2 x
∫
1
6
(
)
=
∫
Check: Let y = sin −1 tanh e 2 x + c , then y ′ =
Hamilton Education Guides
(
)
1
1
1
sec h u du = sin −1 (tanh u ) + c = sin −1 tanh e 2 x + c
6
6
6
1
6 1 − tanh 2 e 2 x
⋅
d
sec h 2 e 2 x
d
⋅ e 2x
tanh e 2 x + 0 =
dx
dx
6 sec h 2 e 2 x
71
Advanced Integration
=
1.4 Integration of Hyperbolic Functions
sec h 2 e 2 x
6 sec h e 2 x
⋅ 2e 2 x =
2e 2 x sec h 2 e 2 x
⋅
6
sec h e 2 x
j. Given ∫ sec h 2 (5 x − 1) dx let u = 5 x − 1 , then
=
e 2x
⋅ sec h e 2 x
3
du d
(5 x − 1) ; du = 5 ; du = 5dx ; dx = du . Therefore,
=
dx
5
dx dx
1
du
1
1
∫ sec h (5x − 1) dx = ∫ sec h 2 u ⋅ 5 = 5 ∫ sec h 2 u du = 5 tanh u + c = 5 tanh (5 x − 1) + c
2
d
1
1
⋅ sec h 2 (5 x − 1) ⋅ (5 x − 1) + 0 = ⋅ sec h 2 (5 x − 1) ⋅ 5
dx
5
5
1
5
Check: Let y = tanh (5 x − 1) + c , then y ′ =
5
⋅ sec h 2 (5 x − 1) = sec h 2 (5 x − 1)
5
du
du
du d
. Therefore,
= 7 ; du = 7 dx ; dx =
=
7x ;
dx
dx dx
7
k. Given ∫ csc h 7 x coth 7 x dx let u = 7 x , then
1
du
1
1
∫ csc h 7 x coth 7 x dx = ∫ csc h u coth u 7 = 7 ∫ csc h u coth u du = − 7 csc h u + c = − 7 csc h 7 x + c
1
7
1
7
Check: Let y = − csc h 7 x + c , then y ′ = − ⋅ − csc h 7 x coth 7 x ⋅
d
1
7 x + 0 = ⋅ csc h 7 x coth 7 x ⋅ 7
dx
7
7
⋅ csc h 7 x coth 7 x = csc h 7 x coth 7 x
7
l. Given ∫ tanh 2 10 x dx let u = 10 x , then
du
du d
du
. Therefore,
= 10 x ;
= 10 ; du = 10 dx ; dx =
dx
dx dx
10
1
du
1
1
∫ tanh 10 x dx = ∫ tanh 2 u ⋅ 10 = 10 ∫ tanh 2 u du = 10 (u − tanh u ) + c = 10 (10 x − tanh 10 x ) + c
2
Check: Let y =
(
1
(10 x − tanh 10 x ) + c , then y ′ = 1 10 − sec h 2 10 x ⋅ d 10 x  + 0 = 1 10 − sec h 2 10 x ⋅10
10 
dx
10
10

(
10 1 − sec h 2 10 x
10
m. ∫ cosh 3
x
dx =
5
) = 1 − sec h 10 x = tanh 10 x
2
2
2 x
x
2 x

x

x
2 x
x
∫ cosh 5 cosh 5 dx = ∫ 1 + sinh 5  cosh 5 dx = ∫  cosh 5 + sinh 5 cosh 5  dx
x
5
= ∫ cosh dx + ∫ sinh 2
x
)
du d x du 1
x
1
x
x
;
= ; du = dx ; dx = 5 du . Therefore,
=
cosh dx let u = , then
dx dx 5 dx 5
5
5
5
5
2 x
x
∫ cosh 5 dx + ∫ sinh 5 cosh 5 dx = ∫ cosh u 5du + ∫ sinh u cosh u 5du = 5 sinh u + 5∫ sinh u cosh u du
2
To solve the second integral let w = sinh u , then
= 5 sinh u + 5∫ w 2 cosh u ⋅
Hamilton Education Guides
2
dw
dw
dw d
thus,
=
= cosh u ; dx =
sinh u ;
cosh u
dx
dx dx
5
x
x
dw
5
= 5 sinh u + 5 w 2 dw = 5 sinh u + w 3 + c = sinh 3 + 5 sinh + c
3
5
5
cosh u
3
∫
72
Advanced Integration
1.4 Integration of Hyperbolic Functions
x
5
5
3
x 1
5 5
x
5
5
3
x 1
5 5
x
5
Check: Let y = 5 sinh + sinh 3 + c , then y ′ = 5 ⋅ cosh ⋅ + ⋅ 3 sinh 2 ⋅ cosh ⋅ + 0
=
x
x
x
x
x
5
x 15
x
x
= cosh 3
⋅ cosh + ⋅ sinh 2 ⋅ cosh = cosh 1 + sinh 2  = cosh ⋅ cosh 2
5
5
5
5
5
5 15
5
5
5
(
)
(
)
n. ∫ sinh 3 x dx = ∫ sinh 2 x sinh x dx = ∫ cosh 2 x − 1 sinh x dx = ∫ cosh 2 x sinh x − sinh x dx
= ∫ cosh 2 x sinh x dx − ∫ sinh x dx = ∫ cosh 2 x sinh x dx − cosh x . To solve the first integral let u = cosh x ,
then
du
du d
du
. Therefore,
=
cosh x ;
= sinh x ; dx =
sinh
x
dx dx
dx
du
1
∫ cosh x sinh x dx − cosh x = ∫ u sinh x ⋅ sinh x − cosh x = ∫ u du − cosh x = 3 u 3 − cosh x + c
2
=
2
2
cosh 3 x
− cosh x + c
3
1
3
Check: Let y = cosh 3 x − cosh + c , then y ′ =
(
)
1
⋅ 3 cosh 2 x ⋅ sinh x − sinh x + 0 = cosh 2 x ⋅ sinh x − sinh x
3
= sinh x cosh 2 x − 1 = sinh x ⋅ sinh 2 x = sinh 3 x
x
2
o. Given ∫ sinh x dx =
1
2
∫ x sinh x dx let u = x and dv = sinh x dx then du = dx and ∫ dv = ∫ sinh x dx
which implies v = cosh x . Using the substitution by parts formula ∫ u dv = u v − ∫ v du we obtain
1
2
1
1
1
2
1
2
1
1
∫ x sinh x dx = 2 x cosh x − 2 ∫ cosh x dx = 2 x cosh x − 2 sinh x + c
Check: Let y = x cosh x − sinh x + c , then y ′ =
−
1
(cosh x + x sinh x ) − 1 cosh x + 0 = cosh x + x sinh x
2
2
2
2
x sinh x
cosh x
1
=
= x sinh x
2
2
2
Example 1.4-3: Evaluate the following indefinite integrals:
a. ∫ tanh 8 x sec h 2 x dx =
b. ∫ tanh 5 (x + 3) sec h 2 (x + 3) dx =
c. ∫ coth 3 x csc h 2 x dx =
d. ∫ coth 5 3x csc h 2 3x dx =
e. ∫ tanh 5 x dx =
f. ∫ 2 tanh
g. ∫ sec h 5 x tanh 5 x dx =
h. ∫ sec h
j. ∫ x 2 coth x 3 dx =
k. ∫ x 2 sec h 5 x 3 dx =
Hamilton Education Guides
x
x
tanh dx =
2
2
x
dx =
3
i. ∫ csc h 2 ( 1 − 2 x ) dx =
l. ∫
sec h x
x
dx =
73
Advanced Integration
1.4 Integration of Hyperbolic Functions
Solutions:
a. ∫ tanh 8 x sec h 2 x dx let u = tanh x , then
du
du d
du
. Thus,
=
tanh x ;
= sec h 2 x ; du = sec h 2 x dx ; dx =
dx dx
dx
sec h 2 x
du
1
1
1
∫ tanh x sec h x dx = ∫ u ⋅ sec h x ⋅ sec h 2 x = ∫ u du = 8 + 1 u 8+1 + c = 9 u 9 + c = 9 tanh x + c
8
2
8
2
1
9
Check: Let y = tanh 9 x + c then y ′ =
8
1
⋅ 9 (tanh x )9−1 ⋅ sec h 2 x + 0 = (tanh x )8 sec h 2 x = tanh 8 x sec h 2 x
9
b. ∫ tanh 5 (x + 3) sec h 2 (x + 3) dx let u = tanh (x + 3) , then
; du = sec h 2 (x + 3) dx ; dx =
= ∫ u 5 du =
du
sec h (x + 3)
2
du
du d
= sec h 2 (x + 3) c ;
tanh (x + 3) ;
=
dx
dx dx
. Thus, ∫ tanh 5 (x + 3) sec h 2 (x + 3) dx = ∫ u 5 ⋅ sec 2 (x + 3) ⋅
du
sec (x + 3)
2
1
1 5+1
1
6
u
+ c = u 6 + c = tanh ( x + 3 ) + c
6
5 +1
6
1
6
Check: Let y = tanh 6 (x + 3) + c then y ′ =
1
⋅ 6 [ tanh (x + 3) ]6−1 ⋅ sec h 2 (x + 3) + 0 = tanh 5 (x + 3) sec h 2 (x + 3)
6
c. Given ∫ coth 3 x csc h 2 x dx let u = coth x , then
; dx = −
9
du
2
csc h x
du
du d
= − csc h 2 x c ; du = − csc h 2 x dx
=
coth x ;
dx
dx dx
. Therefore, ∫ coth 3 x csc h 2 x dx = ∫ u 3 ⋅ csc h 2 x ⋅
−du
2
csc h x
= − ∫ u 3 du =
−1 3+1
u
+c
3 +1
1
4
1
= − u 4 + c = − coth 4 x + c
4
1
4
1
4
Check: Let y = − coth 4 x + c then y ′ = − ⋅ 4(coth x )4−1 ⋅ − csc h 2 x + 0 = coth 3 x csc h 2 x
d. Given ∫ coth 5 3x csc h 2 3x dx let u = coth 3x , then
; dx = −
du
2
3 csc h 3 x
du d
du
= −3 csc h 2 3 x c ; du = −3 csc h 2 3 x dx
=
coth 3 x ;
dx dx
dx
. Therefore, ∫ coth 5 3x csc h 2 3x dx = ∫ u 5 ⋅ csc h 2 3x ⋅
−du
2
3 csc h 3 x
= −
1
u 5 du
3
∫
1
1 1 5+1
1
u
+ c = − u 6 + c = − coth 6 3 x + c
18
3 5 +1
18
= − ⋅
Check: Let y = −
1
1
coth 6 3 x + c then y ′ = − ⋅ 6 (coth 3 x )6−1 ⋅ 3 ⋅ − csc 2 3 x + 0 = coth 5 3 x csc h 2 3 x
18
18
e. Given ∫ tanh 5 x dx let u = 5 x , then
Hamilton Education Guides
du d
du
du
. Therefore,
=
5x ;
= 5 ; du = 5 dx ; dx =
dx dx
dx
5
74
Advanced Integration
1.4 Integration of Hyperbolic Functions
1
du
1
1
∫ tanh 5x dx = ∫ tanh u ⋅ 5 = 5 ∫ tanh u du = 5 ln cosh u + c = 5 ln cosh 5 x + c
1
5
Check: Let y = ln cosh 5 x + c then y ′ =
1
1
5 sinh 5 x
= tanh 5 x
⋅
⋅ sinh 5 x ⋅ 5 + 0 = ⋅
5 cosh 5 x
5 cosh 5 x
x
du 1
x
du d x
;
= ; 3du = dx ; dx = 3du . Therefore,
dx let u = , then
=
3
dx 3
dx dx 3
3
f. Given ∫ 2 tanh
x
x
∫ 2 tanh 3 dx = 2∫ tanh u ⋅ 3du = 6∫ tanh u du = 6 ln cosh u + c = 6 ln cosh 3 + c
x
3
Check: Let y = 6 ln cosh + c , then y ′ = 6 ⋅
x 1
1
6
x
x
= 2 tanh
⋅ sinh ⋅ + 0 = tanh
3 3
3
3
3
cosh 3x
du d
du
du
. Therefore,
=
5x ;
= 5 ; du = 5dx ; dx =
dx dx
dx
5
g. Given ∫ sec h 5 x tanh 5 x dx let u = 5 x , then
1
du
1
1
∫ sec h 5x tanh 5x dx = ∫ sec h u ⋅ tanh u ⋅ 5 = 5 ∫ sec h u tanh u du = − 5 sec h u + c = − 5 sec h 5 x + c
1
5
Check: Let y = − sec h 5 x + c , then y ′ =
h. Given ∫ sec h
x
5 sec h 5 x tanh 5 x
1
= sec h 5 x tanh 5 x
⋅ sec h 5 x tanh 5 x ⋅ 5 + 0 =
5
5
x
x
x
du d x
du 1
;
tanh dx let u = , then
=
= ; 2du = dx ; dx = 2du . Therefore,
2
2
2
dx dx 2
dx 2
x
x
∫ sec h 2 tanh 2 dx = ∫ sec h u ⋅ tanh u ⋅ 2du = 2∫ sec h u tanh u du = − 2 sec h u + c = − 2 sec h 2 + c
x
2
Check: Let y = −2 sec h + c then y ′ = − 2 ⋅ − sec h
i. Given ∫ csc h 2 ( 1 − 2 x ) dx let u = 1− 2 x , then
1
du
x
x 1
2
x
x
x
x
tanh ⋅ + 0 = ⋅ sec h tanh = sec h tanh
2
2 2
2
2
2
2
2
du
du d
du
. Thus,
= 1− 2 x ;
= −2 ; du = −2dx ; dx =
−2
dx dx
dx
1
1
∫ csc h (1 − 2 x ) dx = ∫ csc h u ⋅ − 2 = − 2 ∫ csc h u du = 2 coth u + c = 2 coth ( 1 − 2 x ) + c
2
2
1
2
Check: Let y = coth ( 1 − 2 x ) + c then y ′ =
j. Given ∫ x 2 coth x 3 dx let u = x 3 , then
du
2
1
⋅ − csc h 2 ( 1 − 2 x ) ⋅ −2 + 0 = csc h 2 ( 1 − 2 x )
2
du
du d 3 du
. Therefore,
= 3x 2 ; du = 3 x 2 dx ; dx =
=
x ;
dx
dx dx
3x 2
1
1
1
∫ x coth x dx = ∫ x ⋅ coth u ⋅ 3x 2 = 3 ∫ coth u ⋅ du = 3 ln sinh u + c = 3 ln sinh x + c
2
3
2
1
3
Check: Let y = ln sinh x 3 + c then y ′ =
Hamilton Education Guides
3
3 x 2 cosh x 3
1
1
3
2
⋅
=
⋅
⋅
x
x
cosh
⋅
3
+
0
3 sinh x 3
3 sinh x3
= x 2 coth x 3
75
Advanced Integration
1.4 Integration of Hyperbolic Functions
du
du
du d
. Thus,
5x 3 ;
= 15x 2 ; du = 15 x 2 dx ; dx =
=
dx
dx dx
15x 2
k. Given ∫ x 2 sec h 5 x 3 dx let u = 5x 3 , then
∫ x sec h 5x dx = ∫ x ⋅ sec h u ⋅ 15x 2 = 15 ∫ sec h u ⋅ du = 15 sin −1 (tanh u ) + c = 15 sin ( tanh 5 x )+ c
2
3
Check: Let y =
=
l. Given ∫
∫
15 sec h 5 x
x
x
(
dx =
1
1
−1
3
)
⋅15 x 2 =
3
15 x 2 sec h 2 5 x 3
⋅
15 sec h 5 x 3
1
dx let u = x 2 , then
∫
sec h u
x
Check: Let y = 2 sin
−1
2 sec h 2 x 2
1
sec h x 2
⋅
= x 2 sec h 5 x 3
du d 12 du 1 − 12
1
; dx = 2 x du . Therefore,
=
x ;
= x
=
dx 2
dx dx
2 x
⋅ 2 x du = 2 sec h u ⋅ du = 2 sin −1 (tanh u ) + c
∫
( tanh x )+ c then y ′ =
1
=
1
sec h 2 5 x 3
d
1
d
1
⋅ 5x 3
⋅ tanh 5 x 3 + 0 =
sin −1 tanh 5 x 3 + c then y ′ =
2
3 dx
2
3 dx
15
15 sec h 5 x
15 1 − tanh 5 x
sec h 2 5 x 3
sec h x
sec h x
du
2
1
1
2x 2
=
1
2x 2
⋅
sec h 2 x 2
1
sec h x 2
)
1
1
d
2 sec h 2 x 2 d 12
⋅ tanh x 2 + 0 =
⋅
x
1 dx
1
dx
2 2
2 2
1 − tanh x
sec h x
2
1
2
(
= 2 sin −1 tanh x + c
1
sec h x 2
=
1
x2
=
sec h x
x
Example 1.4-4: Evaluate the following indefinite integrals:
a. ∫ sec h 5 x 2
d. ∫
dx
5
x
3
=
1
dx =
cosh 7 x
g. ∫ 2 sinh
j. ∫ e
3x
3x
csc h
dx =
2
2
cosh x
3 sinh
x
dx =
3
x dx
c. ∫ x 2 sinh x dx =
b. ∫
sinh x
e. ∫
cosh 5 x + sinh 5 x
dx =
sinh 5 x
f. ∫ 
h. ∫ cosh x sec h x + e 3 x dx =
(
i. ∫ e sinh 8 x cosh 8 x dx =
k. ∫ e tanh 5 x sec h 2 5 x dx =
l. ∫ e 3
2
=
)
 1 + cosh x 
 dx =
 sinh x 
1 coth 7 x
csc h 2 7 x dx =
Solutions:
a. Given ∫ sec h 5 x 2
∫
dx
5
sec h x 2
5
x
3
dx
5
x3
= ∫ sec h u ⋅
Hamilton Education Guides
2
let u = x 5 , then
5
5 x 3 du
5
2 x
3
=
2
du d 52 du 2 − 53
5
= x
=
=
x ;
; dx = 5 x 3 du . Therefore,
5 3
dx 5
2
dx dx
5 x
5
5
sec h u ⋅ du = sin −1 (tanh u ) + c
2
2
∫
=
5
5


sin −1  tanh x 2  + c
2


76
Advanced Integration
1.4 Integration of Hyperbolic Functions
5
Check: Let y = sin −1  tanh 5 x 2  + c , then y ′ =
2


2
=
b. Given ∫
x dx
∫ sinh x
5 sec h 2 x 5
2
2 sec h x 5
5x 5
⋅
2
x dx
sinh x
= ∫
2
=
3
10
2
2
⋅
3
sec h 2 x 5
10 x 5
2
=
sec h x 5
sec h x 5
x
du
⋅
sinh u 2 x
1
2
=
∫
du
sinh u
=
=
3
x5
= ∫ x csc h x 2 dx let u = x 2 , then
2
2
2
d
5 sec h 2 x 5 d 52
⋅ tanh x 5 + 0 =
⋅
x
2 dx
2 dx
2 5
2 5
2 1 − tanh x
2 sec h x
5
5
sec h x 2
5
x3
du d 2 du
du
. Thus,
= 2 x ; du = 2 x dx ; dx =
=
x ;
dx
dx dx
2x
x2
1
1
1
u
+c
csc h u du = ln tanh
+ c = ln tanh
2
2
2
2
2
∫
2 x2
2 x2
2
2
2
1
1
2x
1
x2
x 1 − tanh 2
2 x sec h 2
2 x
⋅
sec
h
⋅
+ c , then y ′ = ⋅
Check: Let y = ln tanh
=
= ⋅
⋅
2
2
2
2 tanh x
2 2
2
2
2
4 tanh x
tanh x
=
2
sinh 2 x
2
1−
2
cosh 2 x
x
2
⋅
x2
2
sinh
cosh
=
2
x2
=
2
2
cosh 2 x − sinh 2 x
2
2
2
cosh 2 x
x
2
⋅
x2
2
sinh
cosh
2
x
2
sinh 2 ⋅ x2
=
2
x2
2
1
2 x2
x cosh 2
= ⋅
2 sinh x 2
cosh
2
2
x2
x2
cosh 2
x
= ⋅
2 cosh 2 x 2 ⋅ sinh x 2
2
2
=
x
2
2
2 cosh x2 ⋅ sinh x2
2
x
sinh x 2
c. Given ∫ x 2 sinh x dx let u = x 2 and dv = sinh x dx then du = 2 x dx and ∫ dv = ∫ sinh x dx which implies
implies v = cosh x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
(1 )
∫ x sinh x dx = x ⋅ cosh x − ∫ cosh x ⋅ 2 x dx = x cosh x − 2∫ x cosh x dx
2
2
2
To integrate ∫ x cosh x dx let u = x and dv = cosh x dx then du = dx and ∫ dv = ∫ cosh x dx which
implies v = sinh x . Using the integration by parts formula again we have
(2)
∫ x cosh x dx = x ⋅ sinh x − ∫ sinh x ⋅ dx = x sinh x − cosh x + c
Combining equations ( 1 ) and ( 2 ) together we obtain
∫ x sinh x dx = x cosh x − 2∫ x cosh x dx = x cosh x − 2 (x sinh x − cosh x + c ) = (x + 2)cosh x − 2 x sinh x + c
2
2
2
2
( )
( )
= (x + 2)sinh h x − 2 sinh x = x sinh h x + 2 sinh x − 2 sinh x = x sinh h x
Check: Let y = x 2 + 2 cosh x − 2 x sinh x + c , then y ′ = 2 x cosh x + x 2 + 2 sinh h x − 2 sinh x − 2 x cosh x + 0
2
Hamilton Education Guides
2
2
77
Advanced Integration
d. Given ∫
1.4 Integration of Hyperbolic Functions
1
dx =
cosh 7 x
du
du
d
du
∫ sec h 7 x dx let u = 7 x , then dx = dx 7 x ; dx = 7 ; du = 7dx ; dx = 7 . Thus,
1
1
1
du
1
∫ cosh 7 x dx = ∫ sec h 7 x dx = ∫ sec h u ⋅ 7 = 7 ∫ sec h u du = 7 sin (tanh u ) + c = 7 sin (tanh 7 x ) + c
1
7
Check: Let y = sin −1 (tanh 7 x ) + c , then y ′ =
=
e. Given ∫
−1
−1
sec h 2 7 x
d
d
⋅ 7x
tanh 7 x + 0 =
dx
dx
2
2
7 sec h 7 x
7 1 − tanh 7 x
1
⋅
7 sec h 2 7 x
sec h 2 7 x
1
⋅7 = ⋅
= sec h 7 x =
7 sec h 7 x
7 sec h 7 x
cosh 7 x
cosh 5 x + sinh 5 x
dx =
sinh 5 x
 cosh hx

∫  sinh 5x + 1 dx = ∫ (coth 5x + 1) dx = ∫ coth 5x dx + ∫ dx let u = 5x , then
du d
du
du
. Therefore,
=
5x ;
= 5 ; du = 5dx ; dx =
dx dx
5
dx
∫
cosh 5 x + sinh 5 x
dx =
sinh 5 x
=
1
1
ln sinh u + x + c = ln sinh 5 x + x + c
5
5
 cosh hx

1
5
Check: Let y = ln sinh 5 x + x + c , then y ′ =
f.
 1 + cosh x 
1
du
∫  sinh 5x + 1 dx = ∫ coth 5x dx + ∫ dx = ∫ coth u ⋅ 5x + x = 5 ∫ coth u du + x

5 cosh 5 x
1 cosh 5 x ⋅ 5
+ 1 = coth 5 x + 1
⋅
+1+ 0 = ⋅
5 sinh 5 x
5 sinh 5 x
cosh x 
1
x
∫  sinh x  dx = ∫  sinh x + sinh x  dx = ∫ csc h x dx + ∫ coth x dx = ln tanh 2 + ln sinh x + c
Check: Let y = ln tanh
1
1
d
d
x
⋅ tanh 2x +
⋅ sinh x + 0
+ ln sinh x + c , then y ′ =
x
sinh x dx
dx
2
tanh
2
=
sec h 2 2x cosh x
1
1
2 x 1
+
⋅ sec h 2 ⋅ +
⋅ cosh x =
2 sinh x
2 tanh 2x sinh x
tanh 2x
2 x
1 1 − tanh 2 cosh x
+
= ⋅
2
sinh x
tanh 2x
1
= ⋅
2
1−
sinh 2 x
2
cosh 2 x
sinh x
2
cosh x
2
cosh 2 x − sinh 2 x
2
+
cosh 2 x
cosh x
1
= ⋅
sinh x
2
2
sinh x
2
cosh x
2
2
+
1
cosh 2 x
cosh 2x
cosh x
cosh x
cosh x
1
1
2
+
⋅
+
= ⋅
=
2 sinh 2x
sinh x
2 cosh 2 x ⋅ sinh x sinh x
sinh x
2
2
cosh x
2
cosh x
cosh x
cosh x
1+ cosh x
1
1
1
=
=
=
=
+
+
+
sinh x sinh x
sinh x
sinh 2 ⋅ x sinh x
2 cosh x ⋅ sinh x sinh x
2
g.
∫
2 sinh
2
2
sinh 32x
3x
3x
3x
1
dx = 2
csc h
dx = 2 sinh ⋅
dx = 2 dx = 2 x + c
2 sinh 3 x
2
2
sinh 32x
2
Hamilton Education Guides
∫
∫
∫
78
2
Advanced Integration
1.4 Integration of Hyperbolic Functions
Check: Let y = 2 x + c , then y ′ = 2 ⋅ x1−1 + 0 = 2 ⋅ x 0 = 2
h.
∫ ( cosh x sec h x + e ) dx = ∫  cosh x ⋅ cosh x + e  dx = ∫ (1 + e ) dx = ∫ dx + ∫ e dx = x + 3 e

3x
3x 
1
3x
1
3
1
3
1
3x
3x
+c
3
3
Check: Let y = x + e 3 x + c , then y ′ = x1−1 + ⋅ e 3 x ⋅ 3 + 0 = x 0 + ⋅ e 3 x = 1+ e 3 x
du
du d
du
. Thus,
= 8 cosh 8 x ; dx =
=
sinh 8 x ;
8
cosh
8x
dx dx
dx
i. Given ∫ e sinh 8 x cosh 8 x dx let u = sinh 8 x , then
∫
∫
e sinh 8 x cosh 8 x dx =
e u cosh 8 x ⋅
du
8 cosh 8 x
∫
e
1
eu
1 u
1
du =
e du = e u + c = e sinh 8 x + c
8
8
8
8
1 sinh 8 x
8
⋅ cosh 8 x ⋅ 8 + 0 = ⋅ e sinh 8 x cosh 8 x = e sinh 8 x cosh 8 x
⋅e
8
8
3 sinh
x
x
du d
x
x
du 1
3du
dx let u = cosh , then
. Therefore,
=
cosh ;
= sinh ; dx =
3
dx dx
3
3
3
dx 3
sin x
cosh x
3
cosh x
3 sinh
x
dx =
3
Check: Let y = 3e
∫
x 3du
⋅
3 sinh x
e u sinh
3
cosh x
3 +c
= 3∫ e u du = 3e u + c = 3e
, then y ′ = 3e
cosh x
3 ⋅ sinh
k. Given ∫ e tanh 5 x sec h 2 5 x dx let u = tanh 5 x , then
; dx =
∫e
du
5 sec h 2 5 x
tanh 5 x
sec h 2 5 x dx =
1 coth 7 x
u
du
2
csc h 2 7 x dx let u =
7
3
; du = − csc h 2 7 x dx ; dx = −
∫
1 coth 7 x
3
+c
cosh x
x
3 cosh 3x
x
x 1
3 sinh
⋅ +0 = e
⋅ sinh = e
3 3
3
3
3
du d
du
=
tanh 5 x ;
= 5 sec h 2 5 x ; du = 5 sec h 2 5 x dx
dx dx
dx
csc h 2 7 x dx =
3
7
Hamilton Education Guides
1
∫
u
u
1 tanh 5 x
+c
5
1 tanh 5 x
e
⋅ sec h 2 5 x ⋅ 5 + 0 = e tanh 5 x sec h 2 5 x = e tanh 5 x sec h 2 5 x
5
5
7
du d 1
1
du
=
coth 7 x ;
= − csc h 2 7 x
coth 7 x , then
3
dx dx 3
3
dx
3du
7 csc h 2 7 x
e u csc h 2 7 x ⋅
1 coth 7 x
Check: Let y = − e 3
1
∫ e sec h 5x ⋅ 5 sec h 2 5x = 5 ∫ e du = 5 e + c = 5 e
1
5
l. Given ∫ e 3
cosh x
. Therefore,
Check: Let y = e tanh 5 x + c , then y ′ =
e3
∫
1
8
Check: Let y = e sinh 8 x + c , then y ′ =
j. Given ∫ e
= ∫
. Therefore,
−3du
7 csc h 2 7 x
+ c , then y ′ = −
= −
3 1 coth 7 x
3
3 u
+c
e du = − e u + c = − e 3
7
7
7
∫
1 coth 7 x
3 13 coth 7 x 1
e
⋅ − csc h 2 7 x ⋅ 7 + 0 = e 3
csc h 2 7 x
7
3
79
Advanced Integration
•
1.4 Integration of Hyperbolic Functions
To integrate even powers of sinh x and cosh x use the following identities:
sinh 2 x =
cosh 2 x − sinh 2 x = 1
1
( cosh 2 x − 1)
2
cosh 2 x =
1
( cosh 2 x + 1)
2
To integrate odd powers of sin x and cos x use the following equalities:
∫ sinh
2 n +1
∫ cosh
•
•
2 n +1
∫ ( sinh x ) sinh x dx = ∫ ( cosh x − 1) sinh x dx
( let u = cosh x )
∫ ( cosh x ) cosh x dx = ∫ (1 + sinh x ) cosh x dx
( let u = sinh x )
x dx =
∫ sinh
2n
x sinh x dx =
x dx =
∫ cosh
x cosh x dx =
2n
n
2
2
n
To integrate tanh n x , set
sinh x sinh y =
1
[ cosh (x + y ) − cosh (x − y ) ]
2
cosh x cosh y =
1
[ cosh (x + y ) + cosh (x − y ) ]
2
sinh x cosh y =
1
[ sinh (x + y ) + sinh (x − y ) ]
2
(
) = tanh
n−2
x − tanh n − 2 x sec h 2 x
To integrate coth n x , set
(
) = coth
n−2
x + coth n − 2 x csc h 2 x
coth n x = coth n − 2 x coth 2 x = coth n − 2 x 1 + csc h 2 x
•
n
2
To integrate products of sinh x , sinh y , cosh x , and cosh y use the identities below:
tanh n x = tanh n − 2 x tanh 2 x = tanh n − 2 x 1 − sec h 2 x
•
n
2
To integrate sec h n x
For even powers, set
(
sec h n x = sec h n − 2 x sec h 2 x = 1 − tanh 2 x
)
n−2
2 sec h 2 x
For odd powers change the integrand to a product of even and odd functions, i.e., write
∫ sec h x dx as ∫ sec h x sec h x dx (see Example 1.4-6, problem letter h).
3
•
2
To integrate csc h n x
For even powers, set
(
)
csc h n x = csc h n − 2 x csc h 2 x = coth 2 x − 1
n−2
2 csc h 2 x
For odd powers change the integrand to a product of even and odd functions, i.e., write
∫ csc h x dx as ∫ csc h x csc h x dx (see Example 1.4-6, problem letter i).
3
2
In the following examples we use the above general rules in order to solve integral of products
and powers of hyperbolic functions:
Example 1.4-5: Evaluate the following indefinite integrals:
a. ∫ sinh 5 x cosh 7 x dx =
b. ∫ sinh x cosh x dx =
c. ∫ cosh 3x cosh 2 x dx =
d. ∫ sinh 3x sinh 5 x dx =
e. ∫ cosh 3x cosh 5 x dx =
f. ∫ sinh 5 x dx =
Hamilton Education Guides
80
Advanced Integration
1.4 Integration of Hyperbolic Functions
g. ∫ sinh 3 x dx =
h. ∫ cosh 5 x dx =
i. ∫ tanh 4 x dx =
j. ∫ sinh 7 x dx =
k. ∫ sec h 4 x dx =
l. ∫ cosh 3 x dx =
Solutions:
a.
1
1
=
1
2
=
1
1
cosh 12 x − cosh 2 x + c
24
4
=
1 1
1
1
1
1
12
2
⋅12 sinh 12 x − ⋅ 2 sinh 2 x + 0 =
cosh 12 x − cosh 2 x + c , then y ′ =
sinh 12 x − sinh 2 x
24
24
4
4
24
4
1
1
1
1
1
1
sinh 12 x − sinh 2 x = sinh 12 x + sinh (− 2 x ) = sinh (5 + 7 )x + sinh (5 − 7 )x = sinh 5 x cosh 7 x
2
2
2
2
2
2
1
1
1
1
1 1
1
sinh 2 x dx = ⋅ cosh 2 x = cosh 2 x + c
4
2
2 2
∫
1
4
Check: Let y = cosh 2 x + c , then y ′ =
1
1
1
⋅ 2 sinh 2 x + 0 = sinh 2 x = ⋅ 2 sinh x cosh x = sinh x cosh x
4
2
2
1
1
1
∫ cosh 3x cosh 2 x dx = ∫ 2 [ cosh (3 + 2)x + cosh (3 − 2)x ] dx = ∫ 2 (cosh 5x + cosh x ) dx = 2 ∫ cosh 5x dx
+
1
1
1
1 1
1
sinh 5 x + sinh x + c
cosh x dx = ⋅ sinh 5 x + sinh x + c =
10
2
2
2 5
2
∫
Check: Let y =
=
d.
1 1
∫ sinh x cosh x dx = ∫ 2 [ sinh ( 1 + 1)x + sinh ( 1 − 1)x ] dx = ∫ 2 [ sinh (2 x ) + sinh (0 x ) ] dx = 2 ∫ (sinh 2 x + 0) dx
=
c.
1
∫ (sinh 12 x − sinh 2 x ) dx = 2 ∫ sinh 12 x dx − 2 ∫ sinh 2 x dx = 2 ⋅ 12 cosh 12 x − 2 ⋅ 2 cosh 2 x + c
Check: Let y =
b.
1
∫ sinh 5x cosh 7 x dx = ∫ 2 [ sinh (5 + 7)x + sinh (5 − 7)x ] dx = ∫ 2 [ sinh ( 12 x ) + sinh (− 2 x ) ] dx
1
1
5
1
1
1
⋅ 5 cosh 5 x + ⋅ cosh x + 0 =
cosh 5 x + cosh x
sinh 5 x + sinh x + c , then y ′ =
10
2
10
2
10
2
1
1
1
1
cosh 5 x + cosh x = cosh ( 3 + 2 )x + cosh ( 3 − 2 )x = cosh 3 x cosh 2 x
2
2
2
2
1
1
∫ sinh 3x sinh 5x dx = ∫ 2 [ cosh (3 + 5)x − cosh (3 − 5)x ] dx = ∫ 2 [ cosh (8x ) − cosh (− 2 x ) ] dx
=
1
2
=
1
1
sinh 8 x − sinh 2 x + c
16
4
1
1
1 1
1 1
∫ (cosh 8x − cosh 2 x ) dx = 2 ∫ cosh 8x dx − 2 ∫ cosh 2 x dx = 2 ⋅ 8 sinh 8x − 2 ⋅ 2 sinh 2 x + c
Hamilton Education Guides
81
Advanced Integration
Check: Let y =
=
e.
1.4 Integration of Hyperbolic Functions
1
1
1
1
2
8
sinh 8 x − sinh 2 x + c , then y ′ =
⋅ cosh 8 x ⋅ 8 − ⋅ cosh 2 x ⋅ 2 + 0 =
cosh 8 x − cosh 2 x
16
4
16
4
4
16
1
1
1
1
cosh 8 x − cosh 2 x = [ cosh (8 x ) − cosh (− 2 x ) ] = [ cosh (3 + 5)x − cosh (3 − 5)x ] = sinh 3 x sinh 5 x
2
2
2
2
1
1
∫ cosh 3x cosh 5x dx = ∫ 2 [ cosh (3 + 5)x + cosh (3 − 5)x ] dx = ∫ 2 [ cosh (8x ) + cosh (− 2 x ) ] dx
1
1
1 1
1 1
cosh 8 x dx +
cosh 2 x dx = ⋅ sinh 8 x + ⋅ sinh 2 x + c
2
2
2 8
2 2
1
2
= ∫ (cosh 8 x + cosh 2 x ) dx =
=
∫
∫
1
1
sinh 8 x + sinh 2 x + c
16
4
Check: Let y =
=
1
1
2
8
1
1
sinh 8 x + sinh 2 x + c , then y ′ =
cosh 8 x + cosh 2 x
⋅ cosh 8 x ⋅ 8 + ⋅ cosh 2 x ⋅ 2 + 0 =
16
4
4
16
4
16
1
1
1
1
cosh 8 x + cosh 2 x = [ cosh (8 x ) + cosh (− 2 x ) ] = [ cosh (3 + 5)x + cosh (3 − 5)x ] = cosh 3 x cosh 5 x
2
2
2
2
(
)
(
2
f. ∫ sinh 5 x dx = ∫ sinh 4 x sinh x dx = ∫ sinh 2 x sinh x dx = ∫ cosh 2 x − 1
) sinh x dx . Let u = cosh x , then
2
du
du
du d
. Therefore,
= sinh x ; du = sinh x dx ; dx =
cosh x ;
=
sinh x
dx
dx dx
∫ sinh x dx = ∫ ( cosh x − 1 ) sinh x dx = ∫ ( u − 1 ) sinh x dx = ∫ ( u − 2u + 1)sinh x ⋅ sinh x
5
2
2
)
(
= ∫ u 4 − 2u 2 + 1 du = ∫ u 4 du − 2∫ u 2 du + ∫ du =
1
5
2
2
4
du
2
1
2
1 5 2 3
5
3
u − u + u + c = cosh x − cosh x + cosh x + c
5
3
5
3
2
3
Check: Let y = cosh 5 x − cosh 3 x + cosh x + c , then y ′ =
(
1
2
⋅ 5 cosh 4 x ⋅ sinh x − ⋅ 3 cosh 2 x ⋅ sinh x + sinh x
5
3
(
)
= sinh x cosh 4 x − 2 sinh x cosh 2 x + sinh x = sinh x cosh 4 x − 2 cosh 2 x + 1 = sinh x cosh 2 x − 1
(
= sinh x sinh 2 x
2
) = sinh x sinh x = sinh x
2
4
5
(
)
g. ∫ sinh 3 x dx = ∫ sinh 2 x sinh x dx = ∫ cosh 2 x − 1 sinh x dx . Let u = cosh x , then
;
)
du d
=
cosh x
dx dx
du
du
. Therefore,
= sinh x ; du = sinh x dx ; dx =
sinh x
dx
∫ sinh x dx = ∫ ( cosh x − 1 )sinh x dx = ∫ ( u − 1 ) sinh x ⋅ sinh x = ∫ ( u − 1 ) du = ∫ u du − ∫ du
3
=
2
2
du
2
2
1
1 3
u − u + c = cosh 3 x − cosh x + c
3
3
Hamilton Education Guides
82
Advanced Integration
1.4 Integration of Hyperbolic Functions
1
3
Check: Let y = cosh 3 x − cosh x + c , then y ′ =
(
)
1
⋅ 3 cosh 2 x ⋅ sinh x − sinh x + 0 = sinh x cosh 2 x − sinh x
3
= sinh x cosh 2 x − 1 = sinh x sinh 2 x = sinh 3 x
(
)
(
2
)
2
h. ∫ cosh 5 x dx = ∫ cosh 4 x cosh x dx = ∫ cosh 2 x cosh x dx = ∫ 1 + sinh 2 x cos x dx . Let u = sinh x ,
then
du
du
du d
. Therefore,
= cosh x ; du = cosh x dx ; dx =
=
sinh x ;
cosh x
dx
dx dx
∫ cosh x dx = ∫ (1 + sinh 2 x ) cosh x dx = ∫ (1 + u ) cosh x ⋅ cosh x = ∫ (1 + u 2 ) du = ∫ ( u + 2u + 1)du
= ∫ u 4 du + 2∫ u 2 du + ∫ du =
du
2 2
2
5
4
2
1
2
1 5 2 3
u + u + u + c = sinh 5 x + sinh 3 x + sinh x + c
5
3
5
3
2
3
1
5
2
Check: Let y = sinh 5 x + sinh 3 x + sinh x + c , then y ′ =
(
6
5
sinh 4 x ⋅ cosh x + sinh 2 x ⋅ cosh x + cosh x + 0
3
5
)
(
= cosh x sinh 4 x + 2 cosh x sinh 2 x + cosh x = cosh x sinh 4 x + 2 sinh 2 x + 1 = cosh x 1 + sinh 2 x
(
)
2
) = cosh x cosh x = cosh x
i. ∫ tanh 4 x dx = ∫ tanh 2 x tanh 2 x dx = ∫ tanh 2 x (1 − sec h 2 x ) dx = ∫ tanh 2 x dx − ∫ tanh 2 x sec h 2 x dx
= ∫ (1 − sec h 2 x ) dx − ∫ tanh 2 x sec h 2 x dx = − ∫ tanh 2 x sec h 2 x dx − ∫ sec h 2 x dx + ∫ dx . To solve the first
= cosh x cosh 2 x
2
4
du
du d
du
. Thus,
= sec h 2 x ; du = sec h 2 x dx ; dx =
tanh x ;
=
dx
dx dx
sec h 2 x
integral let u = tanh x , then
∫
5
∫
− tanh 2 x sec h 2 x dx = − u 2 sec h 2 x ⋅
du
2
sec h x
1
3
1
3
= − ∫ u 2 du = − u 3 = − tanh 3 x . Therefore,
1
∫ tanh x dx = ∫ tanh x tanh x dx = − ∫ tanh x sec h x dx − ∫ sec h x dx + ∫ dx = − 3 tanh x − tanh x + x + c
4
2
2
2
1
3
2
2
3
Check: Let y = − tanh 3 x − tanh x + x + c , then y ′ = − tanh 2 x sec h 2 x − sec h 2 x + 1
(
)
)(
(
)
= sec h 2 x − tanh 2 x − 1 + 1 = 1 − tanh 2 x − tanh 2 x − 1 + 1 = − tanh 2 x − 1 + tanh 4 x + tanh 2 x + 1
(
)
= − tanh 2 x + tanh 2 x + (− 1 + 1) + tanh 4 x = tanh 4 x
(
)
(
3
j. ∫ sinh 7 x dx = ∫ sinh 6 x sinh x dx = ∫ sinh 2 x sinh x dx = ∫ cosh 2 x − 1
) sinh x dx . Let u = cosh x , then
3
du
du d
du
. Therefore,
= sinh x ; du = sinh x dx ; dx =
=
cosh x ;
sinh x
dx dx
dx
∫ sinh x dx = ∫ ( cosh x − 1 ) sinh x dx = ∫ ( u − 1 ) sinh x dx = ∫ ( u − 3u + 3u − 1 )sinh x ⋅ sinh x
7
Hamilton Education Guides
2
3
2
3
6
4
2
du
83
Advanced Integration
1.4 Integration of Hyperbolic Functions
(
)
= ∫ u 6 − 3u 4 + 3u 2 − 1 du = ∫ u 6 du − 3∫ u 4 du + 3∫ u 2 du − ∫ du =
=
1
1
1 7
u − 3⋅ u 5 + 3⋅ u 3 − u + c
3
5
7
3
1
cosh 7 x − cosh 5 x + cosh 3 x − cosh x + c
5
7
3
5
1
7
Check: Let y = cosh 7 x − cosh 5 x + cosh 3 x − cosh x + c , then y ′ =
15
7
cosh 6 x ⋅ sinh x − cosh 4 x ⋅ sinh x
5
7
+ 3 cosh 2 x ⋅ sinh x − sinh x + 0 = sinh x cosh 6 x − 3 sinh x cosh 4 x + 3 sinh x cosh 2 x − sinh x
(
)
(
= sinh x cosh 6 x − 3 cosh 4 x + 3 cosh 2 x − 1 = sinh x cosh 2 x − 1
) = sinh x ( sinh x)
3
2
3
= sinh x sinh 6 x = sinh 7 x
(
)
k. ∫ sec h 4 x dx = ∫ sec h 2 x sec h 2 x dx = ∫ sec h 2 x 1 − tanh 2 x dx = ∫ sec h 2 x dx − ∫ tanh 2 x sec h 2 x dx
(
)
= ∫ 1 − tanh 2 x dx − ∫ tanh 2 x sec h 2 x dx = ∫ dx − ∫ tanh 2 x dx − ∫ tanh 2 x sec h 2 x dx . To solve the third
du
du
du d
. Therefore,
= sec h 2 x ; du = sec h 2 x dx ; dx =
tanh x ;
=
dx
dx dx
sec h 2 x
integral let u = tanh x , then
∫
∫
− tanh 2 x sec h 2 x dx = − u 2 sec h 2 x ⋅
du
2
sec h x
1
3
1
3
= − ∫ u 2 du = − u 3 = − tanh 3 x . Thus,
1
∫ sec h x dx = ∫ sec h x sec h x dx = ∫ dx − ∫ tanh x dx − ∫ tanh x sec h x dx = ∫ dx − ∫ tanh 2 x dx − 3 tanh 3 x
4
2
2
2
2
2
1
3
1
1
= x − (x − tanh x ) − tanh 3 x + c = − tanh 3 x + tanh x + (x − x ) + c = − tanh 3 x + tanh x + c
3
3
1
3
3
3
Check: Let y = − tanh 3 x + tanh x + c , then y ′ = − tanh 2 x ⋅ sec h 2 x + sec h 2 x = − tanh 2 x sec h 2 x + sec h 2 x
(
)
= sec h 2 x 1 − tanh 2 x = sec h 2 x sec h 2 x = sec h 4 x
(
)
(
)
l. ∫ cosh 3 x dx = ∫ cosh 2 x cosh x dx = ∫ 1 + sinh 2 x cosh x dx . Let u = sinh x , then
;
du d
=
sinh x
dx dx
du
du
. Therefore,
= cosh x ; du = cosh x dx ; dx =
cosh x
dx
∫ cosh x dx = ∫ (1 + sinh x ) cosh x dx = ∫ (1 + u )cosh x ⋅ cosh x = ∫ (1 + u ) du = ∫ u du + ∫ du
3
=
2
du
2
2
2
1
1 3
u + u + c = sinh 3 x + sinh x + c
3
3
1
3
Check: Let y = sinh 3 x + sinh x + c , then y ′ =
(
)
(
)
1
⋅ 3 sinh 2 x ⋅ cosh x + cosh x + 0 = cosh x sinh 2 x + cosh x
3
= cosh x 1 + sinh 2 x = cosh x cosh 2 x = cosh 3 x
Hamilton Education Guides
84
Advanced Integration
1.4 Integration of Hyperbolic Functions
Example 1.4-6: Evaluate the following indefinite integrals:
a. ∫ cosh 4 x dx =
b. ∫ cosh 2 5 x dx =
c. ∫ sinh 4 x dx =
d. ∫ tanh 3 x dx =
e. ∫ coth 4 x dx =
f. ∫ tanh 6 x dx =
g. ∫ coth 3 x dx =
h. ∫ sec h 3 x dx =
i. ∫ csc h 3 x dx =
Solutions:
a.
1
1
1

2
2
∫ (cosh x ) dx = ∫  2 (cosh 2 x + 1 ) dx = 4 ∫ (cosh 2 x + 1 ) dx = 4 ∫ (1 + cosh 2 x + 2 cosh 2 x )dx
2
2
∫
cosh 4 xdx =
=
2
1
1
x 1
dx +
cosh 2 x dx = +
cosh 2 2 x dx +
4 4
4
4
4
+
3
1
1
x x 1 1
1
1
1
sinh 4 x + sinh 2 x + c
x+
cosh 4 x dx + sinh 2 x = + + ⋅ sinh 4 x + sinh 2 x + c =
8
32
4
4
4
8
4 8 8 4
2
∫
∫
∫
1
1 1
x
1 1
∫ 2 ( 1 + cosh 4 x ) dx + 2 ⋅ 2 sinh 2 x = 4 + 4 ⋅ 2 ∫ dx
∫
Check: Let y =
2
3 4 cosh 4 x 2 cosh 2 x
3 1
1
3x 1
= + cosh 4 x + cosh 2 x
+
+
sinh 4 x + sinh 2 x + c , then y ′ = +
4
32
4
8
8 8
4
8 32
1 1 1
2
2
1
1 1
1 1
2
=  +  + cosh 4 x + cosh 2 x = + ( 1 + cosh 4 x ) + cosh 2 x = + ⋅ ( 1 + cosh 4 x ) + cosh 2 x
4
=
4
8 8
4
8
4
4
(
4 2
4
)
1 1
2
1
1
1

+ ⋅ cosh 2 2 x + cosh 2 x =
1 + cosh 2 2 x + 2 cosh 2 x = ( cosh 2 x + 1 )2 =  ( cosh 2 x + 1 )
4 4
4
4
4
2


(
2
) = cosh x
1
1
b. ∫ cosh 2 5 x dx = ∫ (1 + sinh 2 5 x )dx = ∫ dx + ∫ sinh 2 5 x dx = ∫ dx + ∫ ( cosh 10 x − 1 ) dx = x + ∫ cosh 10 x dx
2
2
= cosh 2 x
−
2
4
x sinh 10 x
1
1 1
x
 1 1
+
+c
dx = x + ⋅ sinh 10 x − + c = x1 −  +
sinh 10 x + c =
2
20
2
2
20
2 10
2


∫
Check: Let y =
1 1
1 1
x sinh 10 x
1 10
⋅ cosh 10 x ⋅10 + 0 = +
+ c , then y ′ = +
+
⋅ cosh 10 x = + cosh 10 x
2 2
2 20
20
2
2 20
1
1
1
1
1
= 1 −  + cosh 10 x = 1 +  cosh 10 x −  = 1 + ( cosh 10 x − 1 ) = 1 + sinh 2 5 x = cosh 2 5 x

c.
2
2
2
2
1
1
1

2
2
∫ (sinh x ) dx = ∫  2 (cosh 2 x − 1 ) dx = 4 ∫ (cosh 2 x − 1 ) dx = 4 ∫ (cosh 2 x − 2 cosh 2 x + 1 )dx
2
2
∫
sinh 4 x dx =
=
1
2
1
1
cosh 2 2 x dx −
cosh 2 x dx +
dx =
4
4
4
4
∫
2
2
∫
Hamilton Education Guides
∫
1
1 1
x
1 1
∫ 2 ( 1 + cosh 4 x ) dx − 2 ⋅ 2 sinh 2 x + 4 = 4 ⋅ 2 ∫ dx
85
Advanced Integration
+
1.4 Integration of Hyperbolic Functions
3
1
1
1
1
1
x
x x 1 1
x+
sinh 4 x − sinh 2 x + c
cosh 4 x dx − sinh 2 x + =  +  + ⋅ sinh 4 x − sinh 2 x + c =
8
32
4
4
4
4
8
8 4 8 4
∫
Check: Let y =
3x 1
1
3 4 cosh 4 x 2 cosh 2 x
2
3 1
= + cosh 4 x − cosh 2 x
+
sinh 4 x − sinh 2 x + c , then y ′ = +
−
8 32
4
4
32
8
4
8 8
1 1 1
2
1 1
1
2
1 1
2
=  +  + cosh 4 x − cosh 2 x = + ( 1 + cosh 4 x ) − cosh 2 x = + ⋅ ( 1 + cosh 4 x ) − cosh 2 x
4
=
8 8
4
4
8
(
1 1
2
1
+ ⋅ cosh 2 2 x − cosh 2 x =
1 + cosh 2 2 x − 2 cosh 2 x
4 4
4
4
(
= sinh 2 x
4
4
4 2
4
) = 14 ( cosh 2 x − 1 ) =  12 ( cosh 2 x − 1 )
2
2
) = sinh x
2
4
(
)
d. ∫ tanh 3 x dx = ∫ tanh 2 x tanh x dx = ∫ 1 − sec h 2 x tanh x dx = − ∫ sec h 2 x tanh x dx + ∫ tanh x dx . To solve
the first integral let u = tanh x , then
∫
∫
− sec h 2 x tanh x dx = − sec h 2 x ⋅ u ⋅
du
du
du d
. Thus,
= sec h 2 x ; du = sec h 2 dx ; dx =
=
tanh x ;
dx
dx dx
sec h 2 x
du
2
sec h x
1
2
1
2
= − ∫ u du = − u 2 = − tan 2 x . Combining the term
∫ tanh x dx = ∫ tanh x tanh x dx = ∫ (1 − sec h x ) tanh x dx = ∫ ( − sec h x tanh x + tanh x ) dx
2
3
2
2
1
2
1
= − ∫ sec h 2 x tanh x dx + ∫ tanh x dx = − tanh 2 x + ∫ tanh x dx = − tanh 2 x + ln cosh x + c
2
1
2
1
2
Check: Let y = − tanh 2 x + ln cosh x + c , then y ′ = − ⋅ 2 tanh x ⋅ sec h 2 x +
= − tanh x sec h 2 x +
(
1
⋅ sinh x + 0
cosh x
sinh x
= − tanh x sec h 2 x + tanh x = tanh x 1 − sec h 2 x
cosh x
(
) = tanh x tanh x = tanh x
2
3
)
e. ∫ coth 4 x dx = ∫ coth 2 x coth 2 x dx = ∫ coth 2 x csc h 2 x + 1 dx = ∫ coth 2 x csc h 2 x dx + ∫ coth 2 x dx
(
)
= ∫ coth 2 x csc h 2 x dx + ∫ csc h 2 x + 1 dx = ∫ coth 2 x csc h 2 x dx + ∫ csc h 2 x dx + ∫ dx .
du
du d
= − csc h 2 x ; du = − csc h 2 x dx
=
coth x ;
dx
dx dx
du
du
; dx = −
. Grouping the terms together we find coth 2 x csc h 2 x dx = u 2 csc h 2 x ⋅ −
2
csc h x
csc h 2 x
To solve the first integral let u = coth x , then
∫
1
3
∫
1
3
= − ∫ u 2 du = − u 3 = − coth 3 x . Therefore,
∫ coth x dx = ∫ coth x coth x dx = ∫ coth x ( csc h x + 1) dx = ∫ coth x csc h x dx + ∫ csc h x dx + ∫ dx
4
2
Hamilton Education Guides
2
2
2
2
2
2
86
Advanced Integration
1.4 Integration of Hyperbolic Functions
1
3
1
= − coth 3 x + ∫ csc h 2 x dx + ∫ dx = − coth3 x − coth x + x + c
3
1
3
3
3
Check: Let y = − coth 3 x − coth x + x + c , then y ′ = − coth 2 x ⋅ − csc h 2 x + csc h 2 x + 1 + 0
(
)
(
)(
)
= coth 2 x csc h 2 x + csc h 2 x + 1 = csc h 2 x coth 2 x + 1 + 1 = coth 2 x − 1 coth 2 x + 1 + 1
= coth 4 x + coth 2 x − coth 2 x − 1 + 1 = coth 4 x
(
)
(
)
f. ∫ tanh 6 x dx = ∫ tanh 4 x tanh 2 x dx = ∫ tanh 4 x 1 − sec h 2 x dx = ∫ tanh 4 x − tanh 4 x sec h 2 x dx =
∫ tanh x dx − ∫ tanh x sec h x dx . In example 1.4-5, problem letter i, we found that
4
4
2
1
∫ tanh x dx = − 3 tanh 3 x − tanh x + x + c . Therefore,
4
∫ tanh x dx = ∫ tanh x tanh x dx = ∫ tanh 4 x (1 − sec h 2 x ) dx = ∫ tanh 4 x dx − ∫ tanh 4 x sec h 2 x dx
6
4
2
1
5
1
3
 1

=  − tanh 3 x − tanh x + x  − ∫ tanh 4 x sec h 2 x dx = − tanh5 x − tanh 3 x − tanh x + x + c
 3

1
5
5
5
1
3
3
3
Check: Let y = − tanh 5 x − tanh 3 x − tanh x + x + c , then y ′ = − tanh 4 x ⋅ sec h 2 x − tanh 2 x ⋅ sec h 2 x
(
)
− sec h 2 x + 1 + 0 = − tanh 4 x sec h 2 x − tanh 2 x sec h 2 x − sec h 2 x + 1 = − sec h 2 x tanh 4 x + tanh 2 x + 1 + 1
(
)(
)
(
)
= − 1 − tanh 2 x tanh 4 x + tanh 2 x + 1 + 1 = − tanh 4 x + tanh 2 x + 1 − tanh 6 x − tanh 4 x − tanh 2 x + 1
= − tanh 4 x − tanh 2 x − 1 + tanh 6 x + tanh 4 x + tanh 2 x + 1 = tanh 6 x
(
)
g. ∫ coth 3 x dx = ∫ coth 2 x coth x dx = ∫ csc h 2 x + 1 coth x dx = ∫ csc h 2 x coth x dx + ∫ coth x dx . To solve the
first integral let u = coth x , then
du
du d
du
. Thus,
=
coth x ;
= − csc h 2 x ; du = − csc h 2 dx ; dx = −
dx dx
dx
csc h 2 x
1
du
1
∫ csc h x coth x dx = ∫ csc h x ⋅ u ⋅ − csc h 2 x = − ∫ u du = − 2 u = − 2 coth x . Combining the term
2
2
2
2
∫ coth x dx = ∫ coth x coth x dx = ∫ ( csc h x + 1) coth x dx = ∫ ( csc h x ⋅ coth x + coth x ) dx
2
3
2
2
1
1
∫ csc h x coth x dx + ∫ coth x dx = − 2 coth x + ∫ coth x dx = − 2 coth x + ln sinh x + c
2
1
2
2
2
1
2
Check: Let y = − coth 2 x + ln sinh x + c , then y ′ = − ⋅ 2 coth x ⋅ − csc h 2 x +
= coth x csc h 2 x +
Hamilton Education Guides
cosh x
sinh x
(
1
⋅ cosh x + 0
sinh x
)
= coth x csc h 2 x + coth x = coth x csc h 2 x + 1 = coth x coth 2 x = coth 3 x
87
Advanced Integration
1.4 Integration of Hyperbolic Functions
)
(
h. ∫ sec h 3 x dx = ∫ sec h 2 x sec h x dx = ∫ 1 − tanh 2 x sec h x dx = − ∫ tanh 2 x sec h x dx + ∫ sec h x dx
= − ∫ tanh x ⋅ tanh x sec h x dx + ∫ sec h x dx . To solve the first integral let u = tanh x and dv = tanh x sec h x dx ,
then du = sec h 2 x dx and ∫ dv = ∫ tanh x sec h x dx which implies v = − sec h x . Using the integration by
parts formula ∫ u dv = u v − ∫ v du we obtain
∫
∫
∫
− tanh 2 x sec h x dx = − tanh x ⋅ tanh x sec h x dx = tanh x sec h x − sec h x sec h 2 x dx = tanh x sec h x
∫
− sec h 3 x dx . Combining the terms we have
∫ sec h x dx = ∫ tanh x ⋅ tanh x sec h x dx − ∫ sec h x dx = tanh x sec h x − ∫ sec h x dx + ∫ sec h x dx . Moving the
3
3
∫ sec h x dx term from the right hand side of the equation to the left hand side we obtain
3
∫ sec h x dx + ∫ sec h x dx = 2∫ sec h x dx = tanh x sec h x + ∫ sec h x dx . Therefore,
3
3
3
∫ sec h x dx = 2 ( tanh x sec h x + ∫ sec h x dx ) = 2 tanh x sec h x + 2 sin ( tanh x ) + c
1
1
3
1
2
1
1
2
Check: Let y = tanh x sec h x + sin −1 (tanh x ) + c , then y ′ =
+
=
=
sec h 2 x
2 1 − tanh 2 x
+0 =
−1
sec h 2 x ⋅ sec h x − sec h x tanh x ⋅ tanh x
2
sec h 3 x − sec h x tanh 2 x
sec h 3 x − sec h x tanh 2 x sec h x
sec h 2 x
+
=
+
2
2
2
2 sec h 2 x
(
sec h 3 x + sec h x 1 − tanh 2 x
sec h 3 x − sec h x tanh 2 x + sec h x
=
2
2
3
3
) = sec h x + sec h x sec h x
3
2
2
3
sec h x + sec h x
2 sec h x
=
= sec h 3 x
2
2
(
)
i. ∫ csc h 3 x dx = ∫ csc h 2 x csc h x dx = ∫ coth 2 x − 1 csc h x dx = ∫ coth 2 x csc h x dx − ∫ csc h x dx
= ∫ coth x ⋅ coth x csc h x dx − ∫ csc h x dx . To solve the first integral let u = coth x and dv = coth x csc h x dx ,
then du = − csc h 2 x dx and ∫ dv = ∫ coth x csc h x dx which implies v = − csc h x . Using the integration by
parts formula ∫ u dv = u v − ∫ v du we obtain
∫ coth x csc h x dx = ∫ coth x ⋅ coth x csc h x dx = coth x ⋅ − csc h x − ∫ csc h x ⋅ csc h x dx = − coth x csc h x
2
2
∫
− csc h 3 x dx . Combining the terms we have
∫ csc h x dx = ∫ coth x ⋅ coth x csc h x dx − ∫ csc h x dx = − coth x csc h x − ∫ csc h x dx − ∫ csc h x dx . Moving
3
Hamilton Education Guides
3
88
Advanced Integration
1.4 Integration of Hyperbolic Functions
the ∫ csc h 3 x dx term from the right hand side of the equation to the left hand side we obtain
∫ csc h x dx + ∫ csc h x dx = 2∫ csc h x dx = − coth x csc h x − ∫ csc h x dx . Therefore,
3
3
3
∫ csc h x dx = 2 ( − coth x csc h x − ∫ csc h x dx ) = − 2 coth x csc h x − 2 ln tanh 2 + c
1
1
3
1
2
(
−
4
(
tanh 2x
sec h 2 2x
(
4 tanh 2x
=
)
sec h 2 2x
csc h 3 x + csc h x coth 2 x
+0 =
−
2
4 tanh x
)
(
2 x
1 1 − tanh 2
1
= ⋅
= ⋅
x
4
4
tanh 2
2
2
)
. The 2nd term is simplified as follows:
cosh 2 x − sinh 2 x
2
cosh 2 x
1
= ⋅
4
)
1
cosh 2 x
2
1
1
2
= ⋅
= ⋅
4 sinh 2x
4
2
sinh x
2
cosh x
2
cosh x
2
cosh
cosh 2
x
2
x
x
⋅ sinh
2
2
csc h x
1
1 1
1
=
=
=
. Therefore,
x
x
x
2 sinh x
2
2 ⋅ sinh 2 ⋅ 2
2 ⋅ 2 cosh ⋅ sinh
2
2
(
=
sinh 2 x
2
cosh 2 x
2
sinh x
2
cosh x
2
1−
sec h 2 2x
csc h 3 x + csc h x coth 2 x
−
2
4 tanh x
=
x
− − csc h x ⋅ csc h x − csc h x coth x ⋅ coth x
x
+ c , then y ′ =
2
2
1
2
Check: Let y = − coth x csc h x − ln tanh
sec h 2 2x
1
2
=
)
csc h 3 x + csc h x coth 2 x csc h x
−
2
2
(
) = csc h x + csc h x csc h x
csc h 3 x + csc h x coth 2 x − csc h x
csc h 3 x + csc h x coth 2 x − 1
=
2
2
3
3
3
2
2
3
sec h x + sec h x
2 csc h x
=
= csc h 3 x
2
2
Example 1.4-7: Evaluate the following indefinite integrals:
a. ∫ sinh 2 x cosh 2 x dx =
b. ∫ sinh 2 x cosh 5 x dx =
c. ∫ sinh 4 x cosh 2 x dx =
d. ∫ sinh 3 x cosh 2 x dx =
e. ∫ tanh 5 x sec h 4 x dx =
f. ∫ tanh 3 x sec h 3 x dx =
g. ∫ coth 2 x csc h 2 x dx =
h. ∫ coth 3 x csc h 3 x dx =
i. ∫ coth 3 x csc h 4 x dx =
Solutions:
a.
1
1
1
1
1
1
∫ sinh x cosh x dx = 4 ∫ sinh 2 x dx = 4 ∫ 2 ( cosh 4 x − 1 ) dx = 8 ∫ ( cosh 4 x − 1) dx = 8 ∫ cosh 4 x dx − 8 ∫ dx
2
=
2
2
1
1
1 1
x
sinh 4 x − x + c
⋅ sinh 4 x − + c =
8
32
8 4
8
Check: Let y =
4 cosh 4 x 1
1
1
1
1
1
sinh 4 x − x + c , then y ′ =
− + 0 = cosh 4 x − = ( cosh 4 x − 1)
8
8
32
8
32
8
8
Hamilton Education Guides
89
Advanced Integration
=
b.
1.4 Integration of Hyperbolic Functions
1 1
1
⋅ ( cosh 4 x − 1) = ⋅ sinh 2 2 x = sinh 2 x cosh 2 x
4 2
4
∫ sinh x cosh x dx = ∫ sinh x cosh x cosh x dx = ∫ sinh 2 x (cosh 2 x ) cosh x dx = ∫ sinh 2 x (1 + sinh 2 x ) cos x dx
2
2
5
(
2
2
4
)
(
)
= ∫ sinh 2 x 1 + sinh 4 x + 2 sinh 2 x cosh x dx = ∫ sinh 2 x cosh x + sinh 6 x cosh x + 2 sinh 4 x cosh x dx
= ∫ sinh 2 x cosh x dx + ∫ sinh 6 x cosh x dx + 2∫ sinh 4 x cosh x dx =
1
3
1
7
3 sinh 2 x cosh x 7 sinh 6 x cosh x
+
7
3
2
5
Check: Let y = sinh 3 x + sinh 7 x + sinh 5 x + c , then y ′ =
+
(
10 sinh 4 x cosh x
= sinh 2 x cosh x + sinh 6 x cosh x + 2 sinh 4 x cosh x = sinh 2 x cosh x 1 + sinh 4 x + 2 sinh 2 x
5
(
= sinh 2 x cosh x 1 + sinh 2 x
c.
2
1
1
sinh 3 x + sinh 7 x + sinh 5 x + c
3
7
5
) = sinh x cosh x ( cosh x) = sinh x cosh x cosh x = sinh x cosh x
2
2
2
2
2
4
2
5
∫ sinh x cosh x dx = ∫ ( sinh x cosh x ) sinh x dx = ∫ ( 12 sinh 2 x ) ⋅ 2 ( cosh 2 x − 1 ) dx = − 8 ∫ sinh 2 2 x dx
4
+
2
2
1
1
sinh 2 2 x cosh 2 x dx = −
8
8
∫
2
2
1
1
1
1
1
1
∫ 2 ( cosh 4 x − 1 ) dx + 8 ∫ sinh 2 x cosh 2 x dx = 16 ∫ dx − 16 ∫ cosh 4 x dx
2
1 1
8 6
1
1
1
1
1
1 1
x−
sinh 3 2 x + c
sinh 4 x +
x − ⋅ sinh 4 x +
sinh 3 2 x + c =
48
64
16
48
16
16 4
Check: Let y =
1 4 cosh 4 x 6 sinh 2 2 x cosh 2 x
1
1
1
−
+
+0
x−
sinh 4 x +
sinh 3 2 x + c , then y ′ =
16
64
48
16
64
48
= + ⋅ sinh 3 2 x =
= −
1
( cosh 4 x − 1) + 1 sinh 2 2 x cos 2 x = − 1 ⋅ 1 ( cosh 4 x − 1) + 1 sinh 2 2 x cosh 2 x
8
8 2
8
16
2
1
2
1
8
1
8
1
2
= − ⋅ sinh 2 2 x + sinh 2 2 x cosh 2 x =  sinh 2 x  ⋅ ( cosh 2 x − 1) = ( sinh x cosh x )2 sinh 2 x

= sinh 2 x cosh 2 x ⋅ sinh 2 x = sinh 4 x cosh 2 x
(
)
(
)
d. ∫ sinh 3 x cosh 2 x dx = ∫ sinh 3 x 1 + sinh 2 x dx = ∫ sinh 3 x + sinh 5 x dx = ∫ sinh 3 x dx + ∫ sinh 5 x dx . In
Example, 1.4-5, problem letters f and g, we found ∫ sinh 5 x dx =
1
1
2
cosh 5 x − cosh 3 x + cosh x + c and
5
3
∫ sinh x dx = 3 cosh x − cosh x + c . Therefore,
3
3
1
1
2

∫ sinh x cosh x dx = ∫ sinh x dx + ∫ sinh x dx = 3 cosh x − cosh x + c +  5 cosh x − 3 cosh x + cosh x + c 
3
2
Hamilton Education Guides
3
5
3
5
3
90
)
Advanced Integration
=
1.4 Integration of Hyperbolic Functions
1
1
1
1
2
cosh 5 x + cosh 3 x − cosh 3 x − cosh x + cosh x + c = cosh 5 x − cosh 3 x + c
3
5
5
3
3
1
5
1
3
Check: Let y = cosh 5 x − cosh 3 x + c , then y ′ =
1
3
⋅ 5 cosh 4 x ⋅ sinh x − ⋅ cosh 2 x ⋅ sinh x + 0
5
3
(
)
= sinh x cosh 4 x − sinh x cosh 2 x = sinh x cosh 2 x cosh 2 x − 1 = sinh x cosh 2 x sinh 2 x = sinh 3 x cosh 2 x
e.
∫ tanh x sec h x dx = ∫ tanh x sec h x sec h x dx = ∫ tanh x (1 − tanh x ) sec h x dx = − ∫ tanh x sec h x dx
5
4
5
∫
+ tanh 5 x sec h 2 x dx = −
2
2
5
2
7
2
2
1
1
tanh8 x + tanh6 x + c
8
6
1
8
1
8
1
6
1
6
Check: Let y = − tanh 8 x + tanh 6 x + c , then y ′ = − ⋅ 8 tanh 7 x ⋅ sec h 2 x + ⋅ 6 tanh 5 x ⋅ sec h 2 x + 0
(
)
= − tanh 7 x sec h 2 x + tanh 5 x sec h 2 x = tanh 5 x sec h 2 x 1 − tanh 2 x = tanh 5 x sec h 2 x sec h 2 x
= tanh 5 x sec h 4 x
(
)
f. ∫ tanh 3 x sec h 3 x dx = ∫ tanh 2 x sec h 2 x ⋅ tanh x sec h x dx = ∫ 1 − sec h 2 x sec h 2 x ⋅ tanh x sec h x dx
= − ∫ sec h 4 x ⋅ tanh x sec h x dx + ∫ sec h 2 x ⋅ tanh x sec h x dx . To solve the first and the second integral
let u = sec h x , then
du
du
du d
. Therefore,
= − sec h x tanh x ; dx = −
=
sec h x ;
sec
h
x tanh x
dx
dx dx
∫ tanh x sec h x dx = ∫ tanh x sec h x ⋅ tanh x sec h x dx = − ∫ sec h x ⋅ tanh x sec h x dx
3
2
3
∫
∫
+ sec h 2 x ⋅ tanh x sec h x dx = − u 4 ⋅ −
=
4
2
tanh x sec h x
tanh x sec h x
du + u 2 ⋅ −
du =
tanh x sec h x
tanh x sec h x
∫
∫ u du − ∫ u du
4
2
1
1
1 5 1 3
3
5
u − u + c = sec h x − sec h x + c
3
5
3
5
1
3
1
5
Check: Let y = sec h 5 x − sec h 3 x + c , then y ′ =
5
3
sec h 4 x ⋅ − sec h x tanh x − sec h 2 x ⋅ − sec h x tanh x
5
3
)
(
= − sec h 4 x ⋅ sec h x tanh x + sec h 2 x ⋅ sec h x tanh x = − sec h 4 x + sec h 2 x sec h x tanh x
(
)
= 1 − sec h 2 x sec h 2 x sec h x tanh x = tanh 2 x sec h 2 x sec h x tanh x = tanh 3 x sec h 3 x
g. Given ∫ coth 2 x csc h 2 x dx let u = coth x , then
∫
coth 2 x csc h 2 x dx =
Hamilton Education Guides
u 2 csc h 2 x
−du
du d
du
. Thus,
=
coth x ;
= − csc h 2 x ; dx =
dx dx
dx
csc h 2 x
1
1
∫ − csc h 2 x du = − ∫ u du = − 3 u 3 + c = − 3 coth x + c
2
3
91
Advanced Integration
1.4 Integration of Hyperbolic Functions
1
3
1
3
Check: Let y = − coth 3 x + c , then y ′ = − ⋅ 3 coth 2 x ⋅ − csc h 2 x + 0 =
h.
3
⋅ coth 2 x csc h 2 x = coth 2 x csc h 2 x
3
∫ coth x csc h x dx = ∫ coth x coth x csc h x dx = ∫ coth x (1 + csc h x ) csc h x dx = ∫ csc h x coth x dx
3
2
3
3
2
2
4
1
1
+ csc h 2 x coth x dx = − csc h 5 x − csc h 3 x + c
5
3
∫
1
5
1
3
5
5
3
3
Check: Let y = − csc h 5 x − csc h 3 x + c , then y ′ = − csc h 4 x ⋅ − csc h x coth x − csc h 2 x ⋅ − csc h x coth x
(
)
= csc h 5 x coth x + csc h 3 x coth x = csc h 3 x coth x csc h 2 x + 1 = csc h 3 x coth x coth 2 x = coth 3 x csc h 3 x
i.
∫ coth x csc h x dx = ∫ coth x csc h x csc h x dx = ∫ coth x ( coth x − 1) csc h x dx = ∫ coth x csc h x dx
3
3
4
2
2
3
2
2
5
2
1
1
− coth 3 x csc h 2 x dx = − coth 6 x + coth 4 x + c
6
4
∫
1
6
1
4
1
6
1
4
Check: Let y = − coth 6 x + coth 4 x + c , then y ′ = − ⋅ 6 coth 5 x ⋅ − csc h 2 x + ⋅ 4 coth 3 x ⋅ − csc h 2 x + 0
(
)
= coth 5 x csc h 2 x − coth 3 x csc h 2 x = coth 3 x csc h 2 x coth 2 x − 1 = coth 3 x csc h 2 x csc h 2 x
= coth 3 x csc h 4 x
Table 1.4-3 provides a summary of the basic integration formulas covered in this manual.
Section 1.4 Practice Problems – Integration of Hyperbolic Functions
1. Evaluate the following integrals:
(
)
a. ∫ cosh 3x dx =
b. ∫ sinh 2 x − e 3 x dx =
d. ∫ x 2 sec h 2 x3dx =
e. ∫ x3 csc h 2 x 4 + 1 dx =
2
3
(
)
g. ∫ cosh 7 ( x + 1) sinh ( x + 1) dx = h. ∫ csc h ( 5 x + 3) coth ( 5 x + 3) dx =
c. ∫ csc h 5 x dx =
(
)
f. ∫ x3 csc h 2 x 4 + 5 dx =
i. ∫ e x +1 sec h e x +1 dx =
2. Evaluate the following integrals:
a. ∫ tanh 5 x sec h 2 x dx =
b. ∫ coth 6 ( x + 1) csc h 2 ( x + 1) dx =
c. ∫ e3 x tanh e3 x dx =
d. ∫ x 3 sec h x 4 + 1 dx =
e. ∫ sec h ( 3x + 2) dx =
f. ∫ ecosh (3 x +5) sinh (3x + 5) dx =
g. ∫ tanh 5 x dx =
h. ∫ coth 5 x dx =
i. ∫ coth 6 x dx =
(
)
Hamilton Education Guides
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Advanced Integration
1.4 Integration of Hyperbolic Functions
Table 1.4-3: Basic Integration Formulas
x n +1
1.
∫ a dx = ax + c
a≠0
2.
∫ x dx = n + 1 + c
3.
∫ a f (x ) dx = a ∫ f (x ) dx
a≠0
4.
∫ [ f (x ) + g (x ) ] dx = ∫ f (x ) dx + ∫ g (x ) dx
5.
∫ sin x dx = − cos x + c
6.
∫ cos x dx = sin x + c
7.
∫ tan x dx = ln sec x + c
8.
∫ cot x dx = ln sin x + c
9.
∫ sec x dx = ln sec x + tan x + c
10.
∫ csc x dx = ln csc x − cot x + c
11.
∫ tan x sec x dx = sec x + c
12.
∫ cot x csc x dx = − csc x + c
13.
∫ sin x dx = 2 −
sin 2 x
+c
4
14.
∫ cos x dx = 2 +
15.
∫ tan x dx = tan x − x + c
16.
∫ tan x dx = tan x − x + c
17.
∫ cot x dx = − cot x − x + c
18.
∫ sec x dx = tan x + c
19.
∫ csc x dx = − cot x + c
20.
∫ x dx = ln x + c
21.
∫ ln xdx = x ln x − x + c
22.
∫
23.
∫ e dx = e + c
24.
∫ a 2 − x 2 dx = a arc sin a + c = a sin
25.
∫ a 2 + x 2 dx = a arc tan a + c = a tan a + c
26.
∫ a 2 − x 2 dx = 2a ln a − x + c
27.
∫ x 2 − a 2 dx = 2a ln x + a + c
28.
∫ x x2 − a 2 dx = a arc sec a + c = a sec a + c
29.
∫ sinh x dx = cosh x + c
30.
∫ cosh x dx = sinh x + c
31.
∫ tanh x dx = ln cosh x + c
32.
∫ coth x dx = ln sinh x + c
33.
∫ sec h x dx = sin (tanh x ) + c
34.
∫ csc h x dx = ln tanh 2 + c
35.
∫ tanh x sec h x dx = − sec h x + c
36.
∫ coth x csc h x dx = − csc h x + c
37.
∫ sinh x dx =
38.
∫ cosh x dx =
39.
∫ tanh x dx = x − tanh x + c
40.
∫ coth x dx = x − coth x + c
41.
∫ sec h x dx = tanh x + c
42.
∫ csc h x dx = − coth x + c
x
2
2
2
2
x
x
1
1
x
1
1
x−a
−1
2
sinh 2 x x
− +c
4
2
2
2
Hamilton Education Guides
1
−1 x
n
x
2
n ≠ −1
sin 2 x
+c
4
2
2
1
a x dx =
ax
+c
ln a
1
1
a 0 and a ≠ 1
1
1
1
1
x
−1 x
a
+c
a+x
1
x
1
−1 x
x
2
sinh 2 x x
+ +c
4
2
2
2
93
Appendix – Exercise Solutions
Section 1.1 Practice Problems – Integration by Parts
1. Evaluate the following integrals using the integration by parts method.
a. Given
1 4x
∫ xe dx let u = x and dv = e dx then du = dx and ∫ dv = ∫ e dx which implies v = 4 e . Using the integration
4x
4x
by parts formula
1
4x
∫ u dv = u v − ∫ v du we obtain
1
1
1
1
1
4x 
∫ xe dx = 4 xe − 4 ∫ e dx = 4 xe − 16 e + c = 4 e  x − 4  + c
4x
Check: Let y =
b. Given
4x
4x
4x
4x
4e 4 x 
1
1
e4 x e4 x
1 1
1 4x 
1

= xe4 x
⋅  x −  + 1 ⋅ e 4 x + 0 = e 4 x ⋅  x −  + e 4 x = xe4 x −
+
e  x −  + c , then y ′ =
4
4
4 
4
4
4 4
4
4


x
∫ 2 cos x dx let u = x and dv = cos x dx then du = dx and ∫ dv = ∫ cos x dx which implies v = sin x . Using the
integration by parts formula
1
1
1
1
x
∫ u dv = u v − ∫ v du we obtain
∫ 2 cos x dx = 2 x ⋅ sin x − 2 ∫ sin x dx = 2 x sin x + 2 cos x + c
Check: Let y =
c. Given
1
1
1
1
1
1
1
1
x sin x + cos x + c , then y ′ = (1 ⋅ sin x + cos x ⋅ x ) − sin x + 0 = sin x + x cos x − sin x = x cos x
2
2
2
2
2
2
2
2
e5x
∫ ( 5 − x )e dx let u = 5 − x and dv = e dx then du = −dx and ∫ dv = ∫ e dx which implies v = 5 . Using
5x
5x
the integration by parts formula
e
5x
∫ ( 5 − x )e dx = (5 − x )
5x
5
Check: Let y = e5 x −
+
∫
5x
∫ u dv = u v − ∫ v du we obtain
1
1 5x
e5 x
dx = e 5 x − xe 5 x +
e +c
5
5
25
(
)
1
1
1
5 5x
1 5x 1 5x
⋅ e + 0 = 5e5 x − e5 x − xe5 x + e5 x
e + c , then y ′ = 5e5 x − e5 x + 5 xe5 x +
xe +
5
25
5
5
25
5
= 5e5 x − xe5 x = (5 − x )e5 x
d. Given
1
∫ x sin 5x dx let u = x and dv = sin 5x dx then du = dx and ∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x .
Using the integration by parts formula
1
∫ u dv = u v − ∫ v du we obtain
1
1
1
∫ x sin 5x dx = x ⋅ − 5 cos 5x + 5 ∫ cos 5x dx = − 5 x cos 5 x + 5 sin 5 x + c
5
1
1
1
1
1
Check: Let y = − x cos 5 x + sin 5 x + c , then y ′ = − (1 ⋅ cos 5 x − sin 5 x ⋅ 5 ⋅ x ) + cos 5 x + 0 = − cos 5 x + x sin 5 x
5
5
5
5
5
5
+
e. Given
1
5
cos 5 x = x sin 5 x = x sin 5 x
5
5
2
3
2
∫ x 3 − x dx let u = x and dv = 3 − x dx then du = dx and ∫ dv = ∫ 3 − x dx which implies v = − 3 (3 − x ) .
Using the integration by parts formula
2
3
2
2
∫ u dv = u v − ∫ v du we obtain
3
2
2
3
2
2
1
∫ x 3 − x dx = x ⋅ − 3 (3 − x ) + ∫ 3 (3 − x ) dx = − 3 x (3 − x ) − 3 ⋅ 1 + 32 (3 − x )
Hamilton Education Guides
3 +1
5
3
2 2
2
2
+c = − x 3− x 2 − ⋅ 3− x 2 +c
3
(
)
3 5
(
)
94
Advanced Integration
= −
Solutions
3
5
2
4
(3 − x ) 2 + c
x (3 − x ) 2 −
3
15
5
3
3
3
1
4
2
4 5
2 3
2
Check: Let y = − x (3 − x ) 2 − (3 − x ) 2 + c , then y ′ = − (3 − x ) 2 + ⋅ x(3 − x ) 2 + ⋅ (3 − x ) 2 + 0
15
3
15 2
3 2
3
1
3
3
1
2
(3 − x ) 2 + x(3 − x ) 2 + 2 (3 − x ) 2 = x(3 − x ) 2 = x 3 − x
3
3
= −
f. Given
3x
3
3 3x
2
3x
3x
∫ u dv = u v − ∫ v du we obtain
integration by parts formula
∫
1
∫ x e dx let u = x and dv = e dx then du = 3x dx and ∫ dv = ∫ e dx which implies v = 3 e . Using the
x3e3 x
1
1 3x
x3e3 x dx = x3 ⋅ e3 x −
− x 2e3 x dx
e ⋅ 3 x 2 dx =
3
3
3
∫
∫
In example 5.1-1, problem letter b, we showed that
∫ x e dx =
3 3x
2 3x
2
3x
2
3x
(
) (
x3e3 x 1 2 3 x 2 3 x 2 3 x
1
1
e + c , then y ′ =
− x e + xe −
3 x 2 ⋅ e3 x + 3e3 x ⋅ x3 − 2 x ⋅ e3 x + 3e3 x ⋅ x 2
27
3
9
3
3
3
(
)
)
2
2
2
2
2
2
⋅ 3e3 x + 0 = x 2e3 x + x3e3 x − xe3 x − x 2e3 x + e3 x + xe3 x − e3 x = x3e3 x
1 ⋅ e3 x + 3e3 x ⋅ x −
3
9
3
9
9
27
+
∫ cos ( ln x ) dx let u = cos ( ln x ) and dv = dx then du =
Using the integration by parts formula
∫ cos ( ln x ) dx = cos ( ln x ) ⋅ x + ∫ x ⋅
To integrate
du =
2 3x
x3e3 x  1 2 3 x 2 3 x 2 3 x 
x 3e 3 x 1 2 3 x 2 3 x
2 3x
−  x e − xe +
e +c =
− x e + xe −
e +c
3
3
9
27
3
3
9
27


Check: Let y =
g. Given
1
∫ x e dx = 3 x e − 9 xe + 27 e + c . Therefore,
− sin ( ln x )
dx and
x
∫ dv = ∫ dx which implies v = x .
∫ u dv = u v − ∫ v du we obtain
sin ( ln x )
dx = x cos ( ln x ) + sin ( ln x ) dx
x
(1 )
∫
∫ sin ( ln x ) dx use the integration by parts formula again, i.e., let u = sin ( ln x ) and dv = dx then
cos ( ln x )
dx and
x
∫ dv = ∫ dx which implies v = x . Therefore,
∫ sin ( ln x ) dx = sin ( ln x ) ⋅ x − ∫ x ⋅
cos ( ln x )
dx = x sin ( ln x ) − cos ( ln x ) dx
x
(2 )
∫
Combining equations ( 1 ) and ( 2 ) together we have
∫ cos ( ln x ) dx = x cos ( ln x ) − ∫ sin ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) − ∫ cos ( ln x ) dx
Taking the integral − cos ( ln x ) dx from the right hand side of the equation to the left hand side
∫
we obtain
∫ cos ( ln x ) dx + ∫ cos ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) Therefore,
2 cos ( ln x ) dx = x cos ( ln x ) + x sin ( ln x ) + c and thus
∫
Check: Let y =
=
h. Given
x
x
∫ cos ( ln x ) dx = 2 cos ( ln x ) + 2 sin ( ln x ) + c
x
x
cos ( ln x ) x sin ( ln x ) sin ( ln x ) x cos ( ln x )
cos ( ln x ) + sin ( ln x ) + c , then y ′ =
+0
−
+
+
2
2
2
2x
2
2x
cos ( ln x ) sin ( ln x ) sin ( ln x ) cos ( ln x )
cos ( ln x ) cos ( ln x )
=
= cos ( ln x )
−
+
+
+
2
2
2
2
2
2
x
∫ 3 tan x dx let u = tan
−1
Hamilton Education Guides
−1
x and dv =
1
x
dx and
dx then du =
3
1+ x 2
x
1
∫ dv = ∫ 3 dx which implies v = 6 x .
95
2
Advanced Integration
Solutions
∫ u dv = u v − ∫ v du we obtain
Using the integration by parts formula
x
x2
1
1
dx
x2
1
1
1 
1

∫ 3 tan 2 x dx = tan x ⋅ 6 − 6 ∫ x ⋅ 1 + x2 = 6 x tan x − 6 ∫ 1 + x2 dx = 6 x tan x − 6 ∫ 1 − 1 + x2  dx
=
−1
−1
2
2
−1
2
−1
1 2 −1
1
1
1
1
1
1
x tan x −
dx +
dx = x 2 tan −1 x − x + tan −1 x + c
6
6
6
6
6 1 + x2
6
∫
∫
1
1 x2
1 1
1
1
1 2
1
+0
− + ⋅
x tan −1 x − x + tan −1 x + c , then y ′ = ⋅ 2 x ⋅ tan −1 x +
6
6
6
6
6 1 + x2 6 6 1 + x2
Check: Let y =
1
1 x2
1
1
1
1 x2 + 1 1
1
1 1
1
1
−1
x
x
⋅ x tan −1 x + ⋅
+
⋅
−
+
⋅
− = x tan −1 x + − = x tan −1 x
tan
=
2
2
2
6 1+ x
3
6 1+ x
6
3
6 1+ x
6
6 6
3
3
i. Given
1
5
∫ ln x dx let u = ln x and dv = dx then du = x5 ⋅ 5x dx = x dx and ∫ dv = ∫ dx which implies v = x . Using the
5
5
integration by parts formula
4
∫ u dv = u v − ∫ v du we obtain
5
∫ ln x dx = ln x ⋅ x − ∫ x ⋅ x dx = x ln x − 5∫ dx = x ln x − 5 x + c
5
5
5
5
5 x5
1


Check: Let y = x ln x5 − 5 x + c , then y ′ = 1 ⋅ ln x5 + 5 ⋅ 5 x 4 ⋅ x  − 5 + 0 = ln x5 + 5 − 5 = ln x5 + 5 − 5 = ln x5
x
x


j. Given
∫x e
− ax
dx let u = x and dv = e − ax dx then du = dx and
integration by parts formula
∫x e
− ax
1
dx which implies v = − e − ax . Using the
a
∫ u dv = u v − ∫ v du we obtain
∫
+
a
a
2
∫
(
)
1 − ax 1 − ax
1
1
1
axe− ax
− 2e
+ c , then y ′ = − 1 ⋅ e − ax − ae − ax ⋅ x − 2 ⋅ −2e − ax + 0 = − e − ax +
xe
a
a
a
a
a
a
⋅ e − ax = −
1 − ax
1
e
+ xe− ax + e − ax = xe− ax
a
a
∫ e sin 3x dx let u = e and dv = sin 3x dx then du = e dx and ∫ dv = ∫ sin 3x dx which implies v = −
x
x
x
Using the integration by parts formula
∫ e sin 3x dx = e ⋅ −
x
x
To integrate
v=
− ax
1
1
1
1 − ax
1
1 − ax
dx = x ⋅ − e − ax +
e dx = − x e − ax +
e dx = − x e − ax − 2 e − ax + c
a
a
a
a
a
a
Check: Let y = −
k. Given
∫ dv = ∫ e
cos 3 x
.
3
∫ u dv = u v − ∫ v du we obtain
1
1 x
cos 3 x
cos 3 x x
e cos 3 x dx
− −
⋅ e dx = − e x cos 3 x +
3
3
3
3
(1 )
∫
∫
∫ e cos 3x dx let u = e and dv = cos 3x dx then du = e dx and ∫ dv = ∫ cos 3x dx which implies
x
x
sin 3 x
. Thus,
3
x
sin 3 x
∫ e cos 3x dx = e ⋅ 3
x
x
−
sin 3 x
1
1
∫ 3 ⋅ e dx = 3 e sin 3x − 3 ∫ e sin 3x dx
x
x
x
(2)
Combining equations (1 ) and ( 2 ) together we obtain:
1
1
1
1
1
∫ e sin 3x dx = − 3 e cos 3x + 3 ∫ e cos 3x dx = − 3 e cos 3x + 9 e sin 3x − 9 ∫ e sin 3x dx
x
Taking the −
x
x
x
x
x
1 x
e sin 3 x dx from the right hand side of the equation to the left hand side we obtain
9
∫
Hamilton Education Guides
96
Advanced Integration
Solutions
1
1
1
10
1
1
∫ e sin 3x dx + 9 ∫ e sin 3x dx = − 3 e cos 3x + 9 e sin 3x which implies 9 ∫ e sin 3x dx = − 3 e cos 3x + 9 e sin 3x
x
and
x
9 1
x
1
x

x
= −
x
x
x
x
3
1
1
3
3 x
1
e cos 3 x + e x sin 3 x , then y ′ = − e x ⋅ cos 3 x + sin 3 x ⋅ 3 ⋅ e x + e x ⋅ sin 3 x + cos 3 x ⋅ 3 ⋅ e x
10
10
10
10
10
10
9
1
3
9 x
1
3 x
10 x
e cos 3 x + e x sin 3 x + e x sin 3 x + e x cos 3 x =
e sin 3 x + e x sin 3 x =
e sin 3 x = e x sin 3 x
10
10
10
10
10
10
10
1
x
x
x
Using the integration by parts formula
∫ u dv = u v − ∫ v du we obtain
1
1
e x sin 5 x 1 x
−
e sin 5 x dx
5
5
∫ e cos 5x dx = e ⋅ 5 sin 5x − 5 ∫ sin 5x ⋅ e dx =
x
x
To integrate
(1 )
∫
x
1
∫ e sin 5x dx let u = e and dv = sin 5x dx then du = e dx and ∫ dv = ∫ sin 5x dx which implies v = − 5 cos 5x .
x
x
1
x
1
∫ e sin 5x dx = e ⋅ − 5 cos 5x + 5 ∫ cos 5x ⋅ e dx = −
x
Thus,
x
∫ e cos 5x dx let u = e and dv = cos 5x dx then du = e dx and ∫ dv = ∫ cos 5x dx which implies v = 5 sin 5x .
l. Given
x
x
Combining equations (1 ) and ( 2 ) together we obtain:
∫ e cos 5x dx =
x
e x cos 5 x 1
+
cos 5 x ⋅ e x dx
5
5
(2)
∫

e x sin 5 x 1  e x cos 5 x 1 x
e x sin 5 x e x cos 5 x
e x sin 5 x 1 x
− −
+
e cos 5 x dx  =
+
−
e sin 5 x dx =

5
5
5
25
5
5 
5
5

∫
∫
1
1
e x cos 5 x dx . Taking the −
e x cos 5 x dx from the right hand side of the equation to the left hand side we obtain
25
25
∫
∫
e x sin 5 x e x cos 5 x
e x sin 5 x e x cos 5 x
26 x
which implies
and
+
+
e cos 5 x dx =
5
25
25
5
25
x
1
e x cos 5 x dx =
25  e x sin 5 x e x cos 5 x 
5 x
1 x
=
+
e sin 5 x +
e cos 5 x

26 
5
25
26
26

∫ e cos 5x dx + 25 ∫ e cos 5x dx =
∫
1
x
∫ e sin 3x dx = 10  − 3 e cos 3x + 9 e sin 3x  = − 10 e cos 3 x + 10 e sin 3 x
Check: Let y = −
−
3
x
Check: Let y =
=
x
∫
1 x
1
5 x
1 x
5 x
5
e sin 5 x +
e cos 5 x , then y ′ =
sin 5 x ⋅ 5 ⋅ e x + 0
cos 5 x ⋅ 5 ⋅ e x +
e ⋅ cos 5 x −
e ⋅ sin 5 x +
26
26
26
26
26
26
5 x
25 x
1 x
5 x
26 x
1 x
25 x
e cos 5 x = e x cos 5 x
e cos 5 x =
e sin 5 x +
e cos 5 x +
e cos 5 x −
e sin 5 x =
e cos 5 x +
26
26
26
26
26
26
26
2. Evaluate the following integrals using the integration by parts method.
∫ x sec x dx let u = x and dv = sec x dx then du = dx and ∫ dv = ∫ sec x dx which implies v = tan x . Using
the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
2
∫ x sec x dx = x ⋅ tan x − ∫ tan x dx = x tan x − ln sec x + c
a. Given
2
2
2
(
) secsecx tanx x + 0 = tan x + x sec x − tan x = x sec x
2
Check: Let y = x tan x − ln sec x + c , then y ′ = 1 ⋅ tan x + sec2 x ⋅ x −
b. Given
∫ arc sin 3 y dy let u = arc sin 3 y and dv = dy then du =
the integration by parts formula
Hamilton Education Guides
3dy
1− 9y
2
and
2
∫ dv = ∫ dy which implies v = y . Using
∫ u dv = u v − ∫ v du we obtain
97
Advanced Integration
Solutions
3dy
∫ arc sin 3 y dy = arc sin 3 y ⋅ y − ∫ y ⋅
∫
3 y dy
∫
To integrate
3 y dy
y dy
1− 9y
(1 )
2
use the substitution method by letting w = 1 − 9 y 2 then
1 − 9 y2
3y
1
dw
−1
1
dw
1
1− 1
1
∫ w ⋅ − 18 y = − 6 ∫ w = − 6 ∫ w dw = − 6 ⋅ 1 − 12 w
=
1 − 9 y2
1− 9y
2
∫
= y arc sin 3 y − 3
(
2
2
dw
dw
. Therefore,
= −18 y and dy = −
dy
18 y
= −
1 2 1
1 1 22−1
= − ⋅ w2
⋅ 2 −1 w
6 1
6
2
)
1
1 1
1
= − w 2 = − 1 − 9 y2 2
3
3
(2 )
Combining equations (1 ) and ( 2 ) together we obtain:
∫
arc sin 3 y dy = y arc sin 3 y −
Check: Let w = y arc sin 3 y +
3y
+
−
1 − 9 y2
c. Given
3 y dy
∫
1 − 9 y2
(
(
)
1
1
1 − 9 y2 2 + c
3
)
1
1
1 − 9 y 2 2 + c , then w ′ = arc sin 3 y +
3
18 y
1
⋅
6
= y arc sin 3 y +
= arc sin 3 y +
1 − 9 y2
3y
3y
1− 9y
3y
−
1 − 9 y2
1 − 9 y2
−
2
1 1
⋅ ⋅
3 2
18 y
1− 9y
2
+ 0 = arc sin 3 y
= arc sin 3 y
dx
∫ arc tan x dx let u = arc tan x and dv = dx then du = 1 + x2 and ∫ dv = ∫ dx which implies v = x . Using the
∫ u dv = u v − ∫ v du we obtain
integration by parts formula
x dx
dx
(1 )
∫ arc tan x dx = arc tan x ⋅ x − ∫ x ⋅ 1 + x2 = x arc tan x − ∫ 1 + x2
x dx
dw
dw
∫ 1 + x2 use the substitution method by letting w = 1 + x then dx = 2 x And dx = 2 x . Therefore,
To integrate
x dx
x dw
2
1
1
1 dw
∫ 1 + x2 = ∫ w ⋅ 2 x = 2 ∫ w = 2 ln w = 2 ln 1 + x
(2)
2
Combining equations (1 ) and ( 2 ) together we obtain:
x dx
1
∫ arc tan x dx = x arc tan x − ∫ 1 + x2 = x arc tan x − 2 ln 1 + x + c
Check: Let y = x arc tan x −
d. Given
2
x
x
x
1 2x
1
−
+ 0 arc tan x +
−
= arc tan x
ln 1 + x 2 + c , then y ′ = arc tan x +
2
2
2
2
2
1+ x
1+ x
1+ x
1 + x2
∫ sin 5x dx = ∫ sin 5x ⋅ sin 5x dx let u = sin 5x and dv = sin 5x dx then du = 10 sin 5x cos 5x dx and
3
2
2
1
∫ dv = ∫ sin 5x dx ∫ dv = ∫ sin x dx which implies v = − 5 cos 5x . Using the integration by parts formula ∫ u dv = u v − ∫ v du
we obtain
∫ sin 5x dx = sin 5x ⋅ −
3
2
1
cos 5 x 1
+
cos 5 x ⋅ 10 sin 5 x cos 5 x dx = − sin 2 5 x cos 5 x + 2 cos 2 5 x sin 5 x dx
5
5
5
∫
∫
(1 )
∫ cos 5x sin 5x dx use the integration by parts method again, i.e., let u = cos 5x and dv = sin 5x then
1
du = −10 sin 5 x cos 5 x dx and ∫ dv = ∫ sin 5 x dx which implies v = − cos 5 x . Therefore,
5
To integrate
2
2
1
1
1
∫ cos 5x sin 5x dx = cos 5x ⋅ − 5 cos 5x − 5 ∫ cos 5x ⋅ 10 sin 5x cos 5x dx = − 5 cos 5x − 2∫ cos 5x sin 5x dx . Taking the
2
Hamilton Education Guides
2
3
2
98
Advanced Integration
Solutions
∫
integral − 2 cos 2 5 x sin 5 x dx from the right hand side of the equation to the left side we obtain
1
1
(2)
∫ cos 5x sin 5x dx + 2∫ cos 5x sin 5x dx = − 5 cos 5x . Therefore, ∫ cos 5x sin 5x dx = − 15 cos 5x
2
2
2
3
3
Combining equations (1 ) and ( 2 ) together we have
1
1
1
∫ sin 5x dx = − 5 sin 5x cos 5x + 2∫ cos 5x sin 5x dx = − 5 sin 5x cos 5x + 2 ⋅ − 15 cos 5x = −
3
−
(
2
2
2
3
sin 2 5 x cos 5 x 2
− cos3 5 x
15
5
)
1
2
1
1
1
2
1

1 − cos 2 5 x cos 5 x − cos3 5 x + c = − cos 5 x +  cos3 5 x − cos3 5 x  + c =
cos 3 5 x − cos 5 x + c
5
15
5
15
5
15
5

Note that another method of solving the above problem is in the following way:
∫ sin 5x dx = ∫ sin 5x ⋅ sin 5x dx = ∫ (1 − cos 5x )⋅ sin 5x dx let u = cos 5x , then dx = −5 sin 5x and dx = − 5 sin 5x .
3
Therefore,
= −
1
5
du
du
2
2
∫ sin 5x dx = ∫ sin 5x ⋅ sin 5x dx = ∫ (1 − cos 5x )⋅ sin 5x dx = ∫ (1 − u )⋅ sin 5x ⋅ − 5 sin 5x
2
3
2
du
2
∫ (1 − u ) du = ∫  5 u − 5  du = 15 u − 5 u + c = 15 cos 5 x − 5 cos 5 x + c
1
2
Check: Let y =
1
1
2
3
1
1
3
1
1
1
1
1
⋅ 3 cos 2 5 x ⋅ − sin 5 x ⋅ 5 + ⋅ 5 sin 5 x + 0 = − cos 2 5 x ⋅ sin 5 x + sin 5 x
cos3 5 x − cos 5 x + c , then y ′ =
15
5
5
15
(
)
= sin 5 x 1 − cos 2 5 x = sin 5 x sin 2 5 x = sin 3 5 x
∫ x cos x dx let u = x and dv = cos x dx then du = 2 x dx and ∫ dv = ∫ cos x dx which implies v = sin x . Using
the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
2
2
2
(1 )
∫ x cos x dx = x ⋅ sin x − ∫ sin x ⋅ 2 x dx = x sin x − 2∫ x sin x dx
To integrate ∫ x sin x dx use the integration by parts formula again, i.e., let u = x and dv = sin x dx then du = dx and
∫ dv = ∫ sin x dx which implies v = − cos x . Using the integration by parts formula ∫ u dv = u v − ∫ v du we obtain
(2)
∫ x sin x dx = x ⋅ − cos x + ∫ cos x ⋅ dx = − x cos x + sin x
2
2
e. Given
Combining equations (1 ) and ( 2 ) together we have
∫ x cos x dx = x sin x − 2∫ x sin x dx = x sin x − 2(− x cos x + sin x ) = x sin x + 2 x cos x − 2 sin x
2
2
2
2
(
)
Check: Let y = x 2 sin x + 2 x cos x − 2 sin x , then y ′ = 2 x sin x + x 2 cos x + 2(cos x − x sin x ) − 2 cos x
2 x sin x + x 2 cos x + 2 cos x − 2 x sin x − 2 cos x = x 2 cos x
f. Given
∫e
−2 x
cos 3 x dx let u = cos 3 x and dv = e −2 x dx then du = −3 sin 3 x dx and
1
v = − e − 2 x . Using the integration by parts formula
2
∫e
−2 x
∫ dv = ∫ e
−2 x
dx which implies
∫ u dv = u v − ∫ v du we obtain
1
3 −2 x
1
3 −2 x
cos 3 x dx = cos 3 x ⋅ − e − 2 x −
⋅ sin 3 x dx = − e − 2 x cos 3 x −
e
e
sin 3 x dx
2
2
2
2
To integrate
∫
∫e
−2 x
∫
(1 )
sin 3 x dx use the integration by parts formula again, i.e., let u = sin 3 x and dv = e −2 x dx then
du = 3 cos 3 x dx and
∫ dv = ∫ e
Hamilton Education Guides
−2 x
1
dx which implies v = − e − 2 x . Therefore,
2
99
Advanced Integration
∫e
−2 x
Solutions
1
1 −2 x
1
3 −2 x
sin 3 x dx = sin 3 x ⋅ − e − 2 x +
e
⋅ 3 cos 3 x dx = − e − 2 x sin 3 x +
e
cos 3 x dx
2
2
2
2
∫
(2)
∫
Combining equations (1 ) and ( 2 ) together we have
∫e
−2 x
cos 3 x dx = −
∫e
Taking the integral
∫e
we obtain
−2 x
1 −2 x
3 −2 x
9 −2 x
1
3
cos 3 x −
sin 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x −
e
e
e
cos 3 x dx
2
4
4
2
2
∫
−2 x
∫
cos 3 x dx from the right hand side of the equation to the left hand side
cos 3 x dx +
1
3
9 −2 x
cos 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x . Therefore,
e
4
2
4
∫
13 − 2 x
3
1
e
cos 3 x dx = − e − 2 x cos 3 x + e − 2 x sin 3 x and thus
4
4
2
∫e
∫
Check: Let y = −
2 −2 x
3 −2 x
cos 3 x +
sin 3 x + c
e
e
13
13
(
)
(
2e −2 x cos 3 x 3e −2 x sin 3 x
3
2
+
+ c , then y ′ = −
− 2e − 2 x cos 3 x − 3e − 2 x sin 3 x +
− 2e − 2 x sin 3 x + 3e − 2 x cos 3 x
13
13
13
13
3
1
∫ x ( 5x − 1) dx let u = x and dv = (5x − 1) dx then du = dx and ∫ dv = ∫ (5x − 1) dx which implies v = 20 (5x − 1) .
3
∫ x ( 5x − 1) dx = x ⋅
Check: Let y =
3
3
Using the integration by parts formula
4
∫ u dv = u v − ∫ v du we obtain
(5 x − 1) 4 − 1 (5 x − 1) 4 dx = x (5 x − 1) 4 − 1 ⋅ 1 (5 x − 1) 4+1 + c = x (5 x − 1) 4 − 1 (5 x − 1) 5 + c
∫
20
20
20
20
20 25
[
500
]
x (5 x − 1) 4
1
(5 x − 1) 5 + c , then y ′ = 1 (5 x − 1) 4 + 20 x (5 x − 1) 3 − 25 (5 x − 1) 4 + 0
−
20
500
500
20
1
(5 x − 1) 4 + x (5 x − 1) 3 − 1 (5 x − 1) 4 = x (5 x − 1) 3
20
20
=
h. Given
cos 3 x dx = −
4 −2 x
6
6
9
4 −2 x
9
cos 3 x + e − 2 x cos 3 x = e −2 x cos 3 x
e
e
cos 3 x + e − 2 x sin 3 x − e − 2 x sin 3 x + e − 2 x cos 3 x =
13
13
13
13
13
13
=
g. Given
−2 x
∫ x csc x dx let u = x and dv = csc x dx then du = dx and ∫ dv = ∫ csc x dx which implies v = − cot x . Using
2
2
the integration by parts formula
2
∫ u dv = u v − ∫ v du we obtain
∫ x csc x dx = x ⋅ − cot x + ∫ cot x dx = − x cot x + ln sin x + c
2
(
x
) cos
+ 0 = − cot x + x csc x + cot x = x csc x
sin x
2
Check: Let y = − x cot x + ln sin x + c , then y ′ = − cot x − x csc2 x +
i. Given
2
∫ 3 cos 5x dx let u = cos 5x and dv = dx then du = −
Using the integration by parts formula
2
2
−1
2
∫
∫
5x
Thus,
−1
5x
1 − 25 x 2
1 − 25 x 2
dx =
Hamilton Education Guides
1 − 25 x 2
dx and
∫ dv = ∫ dx which implies v = x .
∫ u dv = u v − ∫ v du we obtain
∫ 3 cos 5x dx = 3 cos 5x ⋅ x + 3 ∫ x ⋅
To integrate
5
−1
−1
5 dx
1 − 25 x
2
=
2
2
x cos −1 5 x +
3
3
∫
5x
1 − 25 x
2
dx
dx use the substitution method by letting w = 1 − 25 x 2 then
5x
2
dw
1
dw
1
∫ w ⋅ − 50 x = − 10 ∫ w = − 10 ∫
1
1
w2
dw = −
dw
dw
.
= −50 x which implies dx = −
50 x
dx
1
−1
1
1
w 2 dw = − ⋅ 2 w 2 = −
10
10
∫
1 − 25 x 2
5
100
)
Advanced Integration
and
Solutions
2
2
2
cos −15 x dx = x cos −1 5 x +
3
3
3
∫
Check: Let y =
1 − 25 x 2
10 x
10 x
+
3 1 − 25 x
2
3 1 − 25 x
2
=
1
−1
−1
∫ sinh x dx = sinh x ⋅ x − ∫ x ⋅
∫
dw
. Therefore,
2x
(
1
dx
−1
To get the integral of
)
1
= w 2 = 1 + x2 2 =
∫
x dx
1+ x
−
1 − 25 x 2
2
⋅
15
−50 x
2 1 − 25 x 2
+0
dx and
1 + x2
∫ dv = ∫ x dx which implies v = x . Using
∫ u dv = u v − ∫ v du we obtain
the integration by parts formula
−1
5x
2
cos −1 5 x
3
∫ sinh x dx let u = sinh x and dv = dx then du =
j. Given
dx =
∫
2 1 − 25 x 2
2
x cos −1 5 x −
+c
3
15
dx =
2 1 − 25 x 2
2
2
2
x cos −1 5 x −
+ c , then y ′ = cos −1 5 x − ⋅
15
3
3
3
2
cos −1 5 x −
3
=
5x
2
x dx
1 + x2
1+ x
= x sinh −1 x −
2
∫
x dx
1+ x
(1 )
2
use the substitution method by letting w = 1 + x 2 then dw = 2 x dx which implies
=
x
∫ w
⋅
1
dw
=
2
2x
1
∫ w
dw =
1
2
∫
1
1
w2
dw =
−1
1− 1
1
2 12
1 1
2 =
w
w 2 dw = ⋅
w
2
2
2 1− 1
∫
2
(2)
1 + x2
Combining equations (1 ) and ( 2 ) together we have
∫
sinh −1 x dx = x sinh −1 x −
x dx
∫
1 + x2
(
)
(
)
1
= x sinh −1 x − 1 + x 2 2 + c
1
Check: Let y = x sinh −1 x − 1 + x 2 2 + c , then y ′ = sinh −1 x +
x
1+ x
2
2x
2 1+ x
+ 0 = sinh −1 x
2
∫ x sec 10 x dx let u = x and dv = sec 10 x dx then du = dx and ∫ dv = ∫ sec 10 x dx which implies v =
2
2
k. Given
Using the integration by parts formula
tan 10 x
∫ x sec 10 x dx = x ⋅ 10
2
Check: Let y =
=
l. Given
−
−
2
tan 10 x
.
10
∫ u dv = u v − ∫ v du we obtain
1
1
1
tan 10 x dx =
x tan 10 x −
ln sec 10 x + c
10
10
100
∫
1
1 sec 10 x tan 10 x ⋅ 10
1
1
tan 10 x + x sec2 10 x −
+0
x tan 10 x −
ln sec 10 x + c , then y ′ =
10
100
sec 10 x
10
100
1
1
tan 10 x + x sec2 10 x − tan 10 x = x sec2 10 x
10
10
x
x
dx
1
∫ 5 sinh 7 x dx let u = 5 and dv = sinh 7 x dx then du = 5 and ∫ dv = ∫ sinh 7 x dx which implies v = 7 cosh 7 x dx .
Using the integration by parts formula
Hamilton Education Guides
∫ u dv = u v − ∫ v du we obtain
101
Advanced Integration
Solutions
x 1
x
1
1
dx
1
1
1
∫ 5 sinh 7 x dx = 5 ⋅ 7 cosh 7 x − ∫ 7 cosh 7 x ⋅ 5 = 35 x cosh 7 x − 35 ∫ cosh 7 x ⋅ dx = 35 x cosh 7 x − 245 sinh 7 x + c
1
1
1
1
1
1
x cosh 7 x −
cosh 7 x
sinh 7 x + c , then y ′ =
cosh 7 x +
⋅ 7 cosh 7 x + 0 =
⋅ 7 x sinh 7 x −
35
35
245
245
35
35
Check: Let y =
1
1
1
x sinh 7 x − cosh 7 x = x sinh 7 x
5
35
5
+
Section 1.2 Solutions – Integration Using Trigonometric Substitution
Evaluate the following indefinite integrals.
a. Given
dx
∫ 2
16 − x
x
16 − x 2 =
let x = 4 sin t , then dx = 4 cos t dt and
2
(
16 − 16 sin 2 t =
16 1 − sin 2 t
)
16 cos 2 t = 4 cos t . Substituting these values back into the original integral we obtain:
=
∫ 2
x
= −
dx
4 cos t dt
16 − x 2
16 − x 2
4
x
4
1
1 cos t
+c = −
16
16 sin t
(
−2 x
)
16 − x 2
2
= −
16 x
x2
∫
1
1
− x 2 − 16 + x 2
16 x 2 16 − x 2
=
−2 x2
⋅ x − 1 ⋅ 16 − x 2
16 x
+ 0 = − 2 16 − x
2
16
16 x 2 16 − x 2
2
16 − x 2
1
−
16 x
2
− x 2 − 16 − x 2 ⋅ 16 − x 2
16 − x 2
2
= −
16 x
1
=
x 2 16 − x 2
9 − x2 =
dx let x = 3 sin t , then dx = 3 cos t dt and
9 − x2
2
1 4 ⋅ 16 − x 2
16 − x 2
+c
⋅
+c = −
16
4⋅ x
16 x
2
16 − x
+ c , then y ′ = − 2 16 − x
16 x
− x 2 − 16 − x 2
b. Given
+c = −
2
Check: Let y = −
= −
1
4 cos t
∫ (4 sin t ) 2 ⋅ 4 cos t = ∫ 16 sin 2 t ⋅ 4 cos t dt = ∫ 16 sin 2 t dt = 16 ∫ csc t dt = − 16 cot t + c
=
9 − 9 sin 2 t =
9 1 − sin 2 t
(
) = 9 cos t
9

2
= 3 cos t . Substituting these values back into the original integral we obtain:
∫
=
x2
∫
dx =
9 − x2
9 sin 2 t ⋅ 3 cos t dt
=
3 cos t
∫ 9 sin t dt = 9∫
2
1 − cos 2t
9
dt =
2
2
1
∫ (1 − cos 2t ) dt = 2  t − 2 sin 2t  + c
9
x 9 x 9 − x2
9
x x
9
9
+ c = sin −1 −
9 − x2 + c
t − sin t cos t + c = ⋅ sin −1 − ⋅ ⋅
2
3 2 3
3
2
2
2
3 2
9 −1 x x
9
9 − x 2 + c , then y ′ =
sin
−
2
3 2
2
Check: Let y =
=
=
c. Given
∫
9
2
3
9 − x2
9
2 9 − x2
dx
x 9 + 4x2

1  9 − x2
x2
⋅ −
−
3 
2
2 9 − x2

−
(
2 9 − 2x2
4 9 − x2
let x =
Hamilton Education Guides
)=
9
2 9 − x2
1
⋅
2
1 − x9


− 2x
1 1
x
⋅ 
−
9 − x2 +
2
3 2
2 9 − x2


(
)





 2 9 − x2 − 2x2 

 18 − 4 x 2 
9
9
−
−
 =
 =

 4 9 − x2


 4 9 − x2 
2
2
x
2
9
−
2
9
x
−





−
9 − 2x2
2 9 − x2
=
3
3
tan t , then dx = sec2 t dt and
2
2
9 − 9 + 2x2
2 9 − x2
9 + 4x2 =
=
2 x2
2 9 − x2
(
9 + 4 32 tan t
=
x2
9 − x2
) 2 = 9 + 4 ⋅ 94 tan 2 t
102
Advanced Integration
=
(
9 + 9 tan 2 t =
=
∫
Solutions
dx
) = 9 sec t = 3sec t . Therefore,
2
3 sec 2 t
2
∫ 32 tan t ⋅ 3sec t
=
x 9 + 4x2
9 1 + tan 2 t
dt =
1
sec2 t
1
1
1 sec t
1 cos t
1
⋅ sec t dt =
dt =
dt =
⋅
dt
3 tan t ⋅ 3 sec t
3 tan t
3 tan t
3 sin t cos t
∫
9 + 4x2
1
1
1
1
1
dt =
csc t dt = ln csc t − cot t + c = ln
3
3
3 sin t
3
∫
∫
2
9 + 4x − 3
1
ln
3
Check: Let y =
2x
8x2
1
= ⋅
3
=
2x
9 + 4x
1
⋅
3
1
= ⋅
3
9+ 4 x
⋅
2
⋅
1
⋅
3
)
8x2 − 2 ⋅ 9 + 4x2 + 6 9 + 4x2
4x2 9 + 4x2
−
∫
1
= ⋅
3
=
9 + 4x2 − 3
3
1
+ c = ln
2x
3
2 9+ 4 x
⋅
2x
+c


⋅ 8x − 2 ⋅  9 + 4 x2 − 3


1
2x ⋅
2
4x2
9 + 4x2 − 3


− 2 ⋅  9 + 4 x2 − 3


(
2x
2x
4x2
9 + 4x2 − 3
9 + 4x2
+ c , then y ′ =
−3
2x
2x
2
∫
∫
2x


8x2 − 2 ⋅  9 + 4 x2 − 3 9 + 4 x2


⋅
9 + 4 x2 − 3
4 x2 9 + 4 x2
2x
1
⋅
3
⋅
8 x 2 − 18 + 8 x 2 + 6 9 + 4 x 2
4x2 9 + 4x2
9 + 4x2 − 3


6  9 + 4 x2 − 3
6
 = 1 ⋅ 2x ⋅

⋅
3
1
− 3 4x2 9 + 4x2
4x2 9 + 4x2
=
12 x
1
=
12 x 2 9 + 4 x 2
x
9 + 4x2
∫ (49 + x2 )2 dx let x = 7 tan t , then dx = 7 sec t dt and 49 + x = 49 + (7 tan t ) = 49 + 49 tan t = 49(1 + tan t )
1
d. Given
2
2
2
2
2
= 49 sec2 t . Substituting these values back into the original integral we obtain
7 sec2 t dt
7 sec2 t dt
1
7
dt
1
1
1
∫ (49 + x2 )2 dx = ∫ ( 49 sec2 t )2 = ∫ 2401sec2 t sec2 t = 2401 ∫ sec2 t = 343 ∫ cos t dt = 343 ⋅ 2 ∫ (1 + cos 2t ) dt
=

1  −1 x
1
1  1

+
=
=
(
)
t
+
sin
t
cos
t
+
c
t
t
+
c
+
sin
2


 tan
686 
7
686
686 
2



+
(49 − x ) = 49 + x + 49 − x = 98 = 1
( 49 + x )
98( 49 + x )
98( 49 + x )
98( 49 + x )
x2
x2 − 1
(
49 + x 2
(
(
)
2
2
2 2
)


1 
7x 
−1 x
 + c
+c =
 tan 7 +
686
49 + x 2 



dx =
1
x
2
x2
2
Hamilton Education Guides
)
(
)
( ) = 1
) 686( 49 + x ) 98( 49 + x )
+
7 49 − x 2
2 2
2
2 2
2 2
∫
dx + 5 x dx . In Example 5.2-1, problem letter e, the solution to the first integral was:
x −1
x2 − 1 +
(
(
)
2
2 2

)
+
(
1 343 − 7 x 2
1
=
⋅
2
2
686 49 + x
98 49 + x 2
1
7
1 343 + 7 x 2 − 14 x 2
=
⋅
⋅
2
2
686 49 + x 2
686
49 + x
x2
7
⋅
49 + x 2
+
∫  x2 − 1 + 5x  dx = ∫
∫
x
1
49
1
1
1 7 49 + x 2 − 2 x ⋅ 7 x
7x 
1  −1 x
=
⋅
⋅
+
⋅
+
 + c then y ′ =
 tan

2
2
2 2
686
686 71 + x  686
7 49 + x 
686 
7 49 + x 2
49 + x

49 

Check: Let y =
e.
2
1
ln x + x 2 − 1 + c . Therefore, combining the two integrals we have
2
103
Advanced Integration
x2
∫
Solutions
∫
dx + 5 x dx =
2
x −1
Check: Let y =
1
x
2
1
x
2
x2 − 1 +
1
5
ln x + x 2 − 1 + x 2 + c
2
2
x2 − 1 +


 1
1
2x2
1
1
5
ln x + x 2 − 1 + x 2 + c , then y ′ =  x 2 − 1 +
+
2
2
2
2
2 x 2 − 1 
x + x2 − 1






2  x + x2 − 1 
2

 5
2x
1 2 x2 − 1 + 2x2 1
1

 + 5x = 1  4x − 2 +
× 1 +
+
x
⋅
2
+
=


2
2
2
 2 x2 − 1  2
2
2 x2 − 1
x + x2 − 1
2 x2 − 1
 2 x −1


(
1 4x2 − 2 + 2
4x2
+ 5x =
+ 5x =
2
2 x2 − 1
4 x2 − 1
=
f. Given
)
x2
x2 − 1


 + 5x
x 2 − 1 
1
+ 5x
∫ x − 25 dx let x = 5 sec t , then dx = 5 sec t tan t dt and x − 25 = 25 sec t − 25 = 25 tan t = 5 tan t . Thus,
2
2
2
2
∫ x − 25 dx = ∫ 5 tan t ⋅ 5 sec t tan t dt = ∫ 25 sec t tan t dt = 25∫ sec t (sec t − 1) dt = 25∫ sec t dt − 25∫ sec t dt
2
2
2
3
=
25
tan t sec t + ln sec t + tan t
2
=

x 2 − 25
x
25  x
− ln +
⋅

5
5
2 5


x 2
25
25
x 2
x + x 2 − 25
x 2 − 25 
+c =
x − 25 − ln
x − 25 − ln x + x 2 − 25
+c =

2
2
5
2
2
5


+
x
25
ln 5 + c =
2
2
25
25
ln x + x 2 − 25 + c Note:
ln 5 is a constant which can be included in the constant c .
2
2
(
Check: Let y =
x 2 − 25 −
) − 25 ln sec t + tan t + c = 252 ( tan t sec t − ln sec t + tan t ) + c

1
2x2
25
x 2
x − 25 − ln x + x 2 − 25 + c , then y ′ =  x 2 − 25 +
2
2
2
2 x 2 − 25


 25
1
⋅
−
2

x + x 2 − 25

2
2


x 2 − 25 + x 2
25
x + x 2 − 25
1
2x
 + 0 = x − 25 + x − 25 ⋅
× 1 +
−
=
⋅


2
 2 x 2 − 25 
2 x 2 − 25
2 x 2 − 25
x + x 2 − 25
x 2 − 25
2 x 2 − 25
=
g. Given
=
2 x 2 − 50
2 x 2 − 25
) = x − 25 = x − 25 × x − 25 = (x − 25) x − 25 = x − 25
(x − 25)
x − 25
2 x − 25
x − 25
x − 25
(
2 x 2 − 25
2
2
2
2
2
2
2
2
2
2
2
∫ 36 − x dx let x = 6 sin t , then dx = 6 cos t dt and 36 − x = 36 − 36 sin t = 36 cos t = 6 cos t . Thus,
2
2
2
2
36
36 
1
36

∫ 36 − x dx = ∫ 6 cos t ⋅ 6 cos t dt = ∫ 36 cos t dt = 2 ∫ (1 + cos 2t ) dt = 2  t + 2 sin 2t  + c = 2 ( t + sin t cos t ) + c
2
2
=


36  −1 x x 36 − x 2 
36 −1 x x 36 − x 2
36 −1 x 36 x 36 − x 2
+c =
sin
+ ⋅
sin
+
+c =
+
+c
sin


2 
6 6
6
2
6
2 ⋅ 36
2
6
2



Check: Let y =
+
36
36 −1 x x 36 − x 2
+
+ c , then y ′ =
sin
2
2
2
6
36 − x 2 − x 2
2 36 − x 2
=
2 36 − x 2
36 − 2 x 2
+
2 36 − x 2
(36 − x ) 36 − x = 36 − x
2
=
36
36 − x 2
Hamilton Education Guides
2
=
1
( )2
1 − 6x
72 − 2 x 2
2 36 − x 2
=
⋅
1 1
+
36 − x 2 +
6 2
(
2 36 − x 2
2 36 − x 2
− 2 x2
2 36 − x 2
) = 36 − x
2
36 − x 2
=
=
36
⋅
2
6
6 36 − x 2
36 − x 2
36 − x 2
⋅
36 − x 2
36 − x 2
2
104
Advanced Integration
∫
h. Given
dx
Solutions
dx
(9 + 36 x ) ∫ ( + x )
=
3
2 2
let x =
3
2 2
9
36
3
1
1
1

tan t = tan t , then dx = sec2 t dt and 9 + 36 x 2 = 9 + 36 ⋅  tan t 
2
6
2
2

(
2
)
1
= 9 + 36 ⋅ tan 2 t = 9 + 9 tan 2 t = 9 1 + tan 2 t = 9 sec2 t . Therefore,
4
∫
=
(
dx
)
=
i. Given
∫
(
∫
=
)
3
9 sec2 2
1
1
sin t + c =
54
54
Check: Let y =
=
1 sec 2 dt
2
∫
=
3
9 + 36 x 2 2
6x
9 + 36 x
∫
9 2 sec
∫2 3
9
x
9 9 + 36 x
2
sec3 t
)
81 9 + 36 x 2 ⋅ 9 + 36 x 2
2
1
dt
(
)
1
dt
)
9 9 + 36 x 2 −324 x 2
⋅x
2 9 + 36 x 2
=
)
1
=
1
9 + 36 x 2 2
3
3
sin t , then dx = cos t dt and
2
2
dx let x =
(
(
(9 + 36 x )⋅ (
dt
1
∫ 2 729 sec t = 2 ∫ 27 sec t = 54 ∫ sec t = 54 ∫ cos t dt
72 x
81 9 + 36 x 2
1
=
1
2
=
+c
+ c , then y ′ =
81
x
1 sec 2 dt
2
1 ⋅ 9 9 + 36 x 2 − 9 ⋅
9 9 + 36 x 2
9 − 4x2
=
2t
+c =
2
x
(
1 sec 2 dt
2
3
2× 3
(
9 + 36 x 2
81 9 + 36 x 2
=
=
)
(
9 + 36 x 2
81 9 + 36 x 2
)
1
(9 + 36 x )
2
9 − 4x2 =
9 − 4 32 sin t
1
2 1+
81+ 324 x 2 −324 x 2
(9 + 36 x )
3
2 2
(
) 2 = 9 − 4 ⋅ 94 sin 2 t
) = 9 cos t = 3cos t . Therefore,
2
9 − 9 sin 2 t =
9 1 − sin 2 t
9 − 4x2
3 cos t 3
1
6 3 cos 2 t
1 − sin 2 t
1
sin 2 t
⋅
cos
t
dt
=
=
=
dt
⋅
dt
−
3
dt = 3
dt
3
3
dt
3 sin t 2
3 2 sin t
sin t
sin t
sin t
sin t
2
x
dx =
∫
∫
∫
∫
∫
∫
3
−
2x
∫
− 3 sin t dt = 3 csc t dt − 3 sin t dt = 3 ln csc t − cot t + 3 cos t + c = 3 ln
= 3 ln
9 − 4x2
3−
+
2x
Check: Let y = 3 ln
3−
−
×
3−

8x2 − 2 ⋅  3 −

⋅
6x
)−
4x
4x2 9 − 4x2
9 − 4x2


3  − 6 9 − 4 x 2 + 18 


4x
−
9 − 4x2 ⋅ 2x 9 − 4x2
4x
9 − 4x2
=
9 − 4x2
x 9 − 4x2
Hamilton Education Guides
9 − 4x2
+ 3⋅
2x
3
+c
9 − 4x2
=
9 − 4 x2
x 9 − 4 x2
2 9− 4 x
⋅
9 − 4x

9 − 4x2  ⋅ 9 − 4x2

−
=
×
4x
9 − 4x2


18  3 − 9 − 4 x 2 



2
2
 3 − 9 − 4x  ⋅ 2x 9 − 4x


9 − 4 x2
=

9 − 4x2 

−
2
2
9 − 4x2
3−
4x
9 − 4x2
4x
9 − 4x2
(9 − 4 x ) 9 − 4 x =
x (9 − 4 x )
2
6x
=
9 − 4x2 ⋅ 4x2 9 − 4x2
9 − 4 x2
2
4x2


6 x  8 x 2 − 6 9 − 4 x 2 + 18 − 8 x 2 

−
=
3−

⋅ 2x − 2 ⋅  3 −

1
2
4x2 9 − 4x2
9 − 4x2
3−
(
3−
9 − 4x2
8x ⋅
2x
9 − 4 x 2 + c , then y ′ = 3 ⋅
+
8x2 − 6 9 − 4x2 + 2 9 − 4x2
=
−
9 − 4x2
=
2 9 − 4x2
∫
9 − 4x2 + c
2x
8x
∫
=
9
x 9 − 4x2
9 − 4x2
x
105
Advanced Integration
Solutions
Section 1.3 Solutions – Integration by Partial Fractions
a. Evaluate the integral
dx
∫ x2 + 5x + 6 .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
Second - Factor the denominator x 2 + 5 x + 6 into (x + 2 ) (x + 3) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once,
the integrand can be represented in the following way:
2
1
=
x + 5x + 6
1
( x + 2 ) ( x + 3)
B
A
+
x+2 x+3
=
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
1
2
x + 5x + 6
=
A (x + 3) + B (x + 2 )
(x + 2) (x + 3)
1 = A (x + 3) + B (x + 2 ) = Ax + 3 A + Bx + 2 B
1 = ( A + B ) x + (3 A + 2 B ) therefore,
A+ B = 0
3 A + 2B = 1
which result in having A = 1 and B = −1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.
dx
A
1
B
1
∫ x2 + 5x + 6 = ∫ x + 2 dx + ∫ x + 3 dx = ∫ x + 2 dx − ∫ x + 3 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
∫ x + 2 dx − ∫ x + 3 dx = ln x + 2 − ln x + 3 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = ln x + 2 − ln x + 3 + c , then y ′ =
b. Evaluate the integral
1
( x + 3) − ( x + 2 ) = x + 3 − x − 2 =
1
1
⋅1 −
⋅1 + 0 =
(x + 2) (x + 3) x 2 + 3x + 2 x + 6 x 2 + 5 x + 6
x+2
x+3
x 2 +1
∫ x 3 − 4 x dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
(
)
Second - Factor the denominator x3 − 4 x into x x 2 − 4 = x(x − 2 ) (x + 2 ) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once,
the integrand can be represented in the following way:
x2 + 1
=
3
x − 4x
x2 + 1
A
B
C
=
+
+
x x−2 x+2
x (x − 2 ) (x + 2 )
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x2 + 1
(
x3 − 4 x
=
A (x − 2 ) (x + 2 ) + Bx (x + 2 ) + Cx (x − 2 )
x (x − 2 ) (x + 2 )
) (
) (
)
x 2 + 1 = A x 2 + 2 x − 2 x − 4 + B x 2 + 2 x + C x 2 − 2 x = Ax 2 − 4 A + Bx 2 + 2 Bx + Cx 2 − 2Cx
x 2 + 1 = ( A + B + C )x 2 + (2 B − 2C )x − 4 A therefore,
A+ B+C =1
Hamilton Education Guides
2 B − 2C = 0
−4 A = 1
106
Advanced Integration
Solutions
which result in having A = −
5
1
5
, B = , and C =
8
4
8
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.
x 2 +1
A
B
1 1
C
5
1
5
1
∫ x 3 − 4 x dx = ∫ x dx + ∫ x − 2 dx + ∫ x + 2 dx = − 4 ∫ x dx + 8 ∫ x − 2 dx + 8 ∫ x + 2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
−
1 1
5
dx +
4 x
8
∫
1
5
5
1
1
5
∫ x − 2 dx + 8 ∫ x + 2 dx = − 4 ln x + 8 ln x − 2 + 8 ln x + 2 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
5
5
1
−2(x − 2 )(x + 2 ) + 5 x(x + 2 ) + 5 x(x − 2 )
1
5
5
+
+
=
Let y = − ln x + ln x − 2 + ln x + 2 + c , then y ′ = −
4
8
8
8 x(x − 2 )(x + 2 )
4 x 8(x − 2 ) 8(x + 2 )
=
− 2 x 2 + 8 + 5 x 2 + 10 x + 5 x 2 − 10 x
(
8x x2 − 4
c. Evaluate the integral
)
=
8x2 + 8
(
8x x2 − 4
=
)
( ) = x +1
x − 4x
8 x (x − 4 )
2
8 x2 + 1
3
2
1
∫ 36 − x 2 dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
Second - Factor the denominator 36 − x 2 into (6 − x )(6 + x ) .
Third - Write the linear factors in partial fraction form. Since each linear factor in the denominator is occurring only once,
the integrand can be represented in the following way:
1
36 − x
2
=
1
(6 − x ) (6 + x )
=
A
B
+
6−x 6+ x
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
1
36 − x
2
=
A (6 + x ) + B (6 − x )
(6 − x ) (6 + x )
1 = A (6 + x ) + B (6 − x ) = 6 A + Ax + 6 B − Bx
1 = ( A − B ) x + (6 A + 6 B ) therefore,
6 A + 6B = 1
which result in having A =
A− B = 0
1
1
, and B =
12
12
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.
1
A
B
1
1
1
1
∫ 36 − x2 dx = ∫ 6 − x dx + ∫ 6 + x dx = 12 ∫ 6 − x dx + 12 ∫ 6 + x dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
1
1
1
1
dx +
dx =
ln 6 − x +
ln 6 + x + c
12 6 − x
12 6 + x
12
12
∫
∫
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
1
6+ x+6− x
1
1
1
1
1
1
=
ln 6 − x + ln 6 + x + c , then y ′ =
⋅
+
⋅
+0 =
12
12
12(6 − x ) (6 + x )
12 6 − x 12 6 + x
36 − x 2
d. Evaluate the integral
x+5
∫ x3 + 2 x2 + x dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
Hamilton Education Guides
107
Advanced Integration
Solutions
(
)
Second - Factor the denominator x3 + 2 x 2 + x into x x 2 + 2 x + 1 = x(x + 1)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand
can be represented in the following way:
x+5
x3 + 2 x 2 + x
=
(
x+5
=
)
2
x x + 2x + 1
x+5
=
x(x + 1)2
A
B
C
+
+
x x + 1 (x + 1)2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
x+5
x3 + 2 x 2 + x
(
A (x + 1)2 + Bx (x + 1) + Cx
=
) (
x(x + 1)(x + 1)2
)
x + 5 = A x 2 + 2 x + 1 + B x 2 + x + Cx = Ax 2 + 2 Ax + A + Bx 2 + Bx + Cx
x + 5 = ( A + B )x 2 + (2 A + B + C )x + A therefore,
2A + B + C = 1
A+ B = 0
A=5
which result in having A = 5 , B = −5 , and C = −4
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.
x+5
A
1
C
B
1
1
∫ x3 + 2 x2 + x dx = ∫ x dx + ∫ x + 1 dx + ∫ (x + 1) 2 dx = 5∫ x dx − 5∫ x + 1 dx − 4∫ (x + 1) 2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
5
1
1
4
1
∫ x dx − 5∫ x + 1 dx − 4∫ (x + 1) 2 dx = 5 ln x − 5 ln x + 1 + x + 1 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = 5 ln x − 5 ln x + 1 +
=
5(x + 1) 2 − 5 x (x + 1) − 4 x
4
1
4
1
+ c , then y ′ = 5 ⋅ − 5 ⋅
+0 =
−
2
x +1
x
x + 1 (x + 1)
x (x + 1)2
5 x 2 + 10 x + 5 − 5 x 2 − 5 x − 4 x
3
2
x + 2x + x
e. Evaluate the integral
=
x+5
x + 2x2 + x
3
1
∫ x3 − 2 x2 + x dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
Second – Factor the denominator x3 − 2 x 2 + x into x(x − 1)2 .
Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand
can be represented in the following way:
1
x(x − 1)2
=
A
B
C
+
+
x x − 1 (x − 1)2
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
1
x3 − 2 x 2 + x
(
) (
A (x − 1)2 + Bx (x − 1) + Cx
=
)
x(x − 1)2
1 = A x 2 − 2 x + 1 + B x 2 − x + Cx = Ax 2 − 2 Ax + A + Bx 2 − Bx + Cx
1 = ( A + B )x 2 + (− 2 A − B + C )x + A therefore,
A+ B = 0
−2 A − B + C = 0
A =1
which result in having A = 1 , B = −1 , and C = 1
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.
Hamilton Education Guides
108
Advanced Integration
dx
Solutions
A
B
1
C
1
1
∫ x3 − 2 x2 + x = ∫ x dx + ∫ x − 1 dx + ∫ (x − 1)2 dx = ∫ x dx − ∫ x − 1 dx + ∫ (x − 1) 2 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
1
1
∫ x dx − ∫ x − 1 dx + ∫ (x − 1) 2 dx = ln x − ln x − 1 − x − 1 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = ln x − ln x − 1 −
=
x2 − x2 − 2x + 2x + 1
3
=
2
x − 2x + x
f. Evaluate the integral
(x − 1)2 − x (x − 1) + x = x 2 − 2 x + 1 − x 2 + x + x
1
1
1
1
+ c , then y ′ = −
+
+0 =
2
x −1
x x − 1 (x − 1)
x3 − 2 x 2 + x
x(x − 1)2
3
1
x − 2x2 + x
x2 + 3
∫ x2 − 1 dx .
(x − 1)+ 4
x2 + 3
2
 x2 − 1
4


4

∫ x2 − 1 dx = ∫ x2 − 1 dx = ∫  x2 − 1 + x2 − 1  dx = ∫ 1 + x 2 − 1  dx .
First – Rewrite the integral in the following form:
Then, check to see if the integrand of the second integral is a proper or an improper rational fraction. If the integrand is an
improper rational fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and
a proper rational fraction.
Second - Factor the denominator x 2 − 1 into (x − 1) (x + 1) .
Third - Write the linear factors in partial fraction form. Since one of the factors in the denominator is repeated, the integrand
can be represented in the following way:
4
x 2 −1
=
4
=
(x − 1) (x + 1)
B
A
+
x −1 x +1
Fourth - Solve for the constants A and B by equating coefficients of the like powers.
4
x 2 −1
=
A (x + 1) + B (x − 1)
(x − 1) (x + 1)
4 = A (x + 1) + B (x − 1) = Ax + A + Bx − B
4 = ( A + B )x + ( A − B ) therefore,
A− B = 4
A+ B = 0
which result in having A = 2 and B = −2
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants with their specific values.
x2 + 3

4
4

A
B
1
1
∫ x2 − 1 dx = ∫ 1 + x2 − 1  dx = ∫ dx + ∫ x2 − 1 dx = ∫ dx + ∫ x − 1 dx + ∫ x + 1 dx = ∫ dx + 2∫ x − 1 dx − 2∫ x + 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
∫ dx + 2∫ x − 1 dx − 2∫ x + 1 dx = x + 2 ln x − 1 − 2 ln x + 1 + c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y = x + 2 ln x − 1 − 2 ln x + 1 + c , then y ′ = 1 + 2 ⋅
=
(x − 1) (x + 1) + 2(x + 1) − 2(x − 1)
1
1
− 2⋅
+0 =
(x − 1) (x + 1)
x −1
x +1
(x + x − x − 1)+ 2 x + 2 − 2 x + 2 = x − 1 + 4 = x + 3
x −1
x −1
(x + x − x − 1)
2
2
2
2
2
g. Evaluate the integral
2
1
∫ x3 − 1 dx .
Hamilton Education Guides
109
Advanced Integration
Solutions
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
(
)
Second - Factor the denominator x3 − 8 into (x − 1) x 2 + x + 1 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the
integrand can be represented in the following way:
1
3
x −1
1
=
(x − 1) (x 2 + x + 1)
=
A
Bx + C
+ 2
x −1 x + x +1
Fourth - Solve for the constants A , B and C by equating coefficients of the like powers.
1
3
x −1
(
=
(
)
(x − 1) (x + x + 1)
A x 2 + x + 1 + (Bx + C ) (x − 1)
2
)
1 = A x 2 + x + 1 + (Bx + C ) (x − 1) = Ax 2 + Ax + A + Bx 2 − Bx + Cx − C
1 = ( A + B ) x 2 + ( A − B + C )x + ( A − C ) therefore,
A+ B = 0
which result in having A =
A− B+C = 0
A−C =1
1
2
1
, B = − , and C = −
3
3
3
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
∫ x3 − 1
dx =
∫
A
Bx + C
1
1
dx + 2
dx =
dx +
x −1
3 x −1
x + x +1
∫
∫
− 13 x − 23
1
1
1
x+2
∫ x2 + x + 1 dx = 3 ∫ x − 1 dx − 3 ∫ x2 + x + 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous
sections. To solve the second integral let u = x 2 + x + 1 , then
x+2 =
1
(2 x + 1) + 3 . Therefore,
2
2
1
1
1
x+2
1
1
1
dx −
dx =
dx −
3 x −1
3 x2 + x + 1
3 x −1
3
∫
=
=
∫
∫
)
1 2x + 1 + 3
2 dx = 1
2
2
3
∫ x + x +1
∫
1
1
dx −
x −1
6
2x + 1
∫ x2 + x + 1
1
1
1 2 x + 1 du
1
1
1
1
1
1
2x + 1
⋅
−
dx −
dx −
dx −
dx =
2
2
3 x −1
6
2x + 1 2
u
2 x + x +1
3 x −1
6 x + x +1
∫
∫
∫
∫
1
1
1 1
ln x − 1 − ln u − ⋅
tan −1
3
6
2 3
2
=
(
du
du
. Also, x + 2 can be rewritten as
= 2 x + 1 and dx =
2x + 1
dx
∫
2 x +1
2 + c = 1 ln x − 1 − 1 ln x 2 + x + 1 − 1 ⋅
3
2
3
6
2
2
3
dx −
3
1
2
dx
3 x2 + x + 1
∫
1
∫ (x + 1 ) 2 + 3 dx
tan −1
2
4
2(2 x + 1)
2 3
+c
1
1
1
2x + 1
1
1
3
2x + 1
ln x − 1 − ln x 2 + x + 1 −
tan −1
tan −1
+c
+ c = ln x − 1 − ln x 2 + x + 1 −
3
6
3
6
3
3
3
3
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
y′ =
−
1
1
3
2x + 1
ln x − 1 − ln x 2 + x + 1 −
tan −1
+ c , then
3
6
3
3
3
1 1
1
1
− ⋅
⋅ (2 x + 1) −
⋅
⋅
3
3 x − 1 6 x2 + x + 1
1
( 2 ⋅ 3 )− 0 ⋅ (2 x + 1) + 0 = 1 − 2 x + 1
3(x − 1) 6(x + x + 1)

( 3)

1 +  2 x +1
 3 
2
⋅
2
2
1
2x + 1
2
3
3
2 3
6 x 2 + 6 x + 6 − (2 x + 1) ⋅ (3 x − 3)
2
=
=
−
−
⋅
⋅
−
2
2
2
3(x − 1) 6 x 2 + x + 1
3 3 + (2 x + 1)
3
3 + 4x + 4x + 1
3 + (2 x + 1)
18(x − 1) x 2 + x + 1
Hamilton Education Guides
(
)
(
)
110
Advanced Integration
=
Solutions
6 x 2 + 6 x + 6 − 6 x 2 + 6 x − 3x + 3
−
2
9x + 9
=
1
−
x +1
=
1
−
)
18(x − 1) 4(x + x + 1)
2(x − 1) 2(x + x + 1)
1
(x + 1) (x + x + 1) − (x − 1) = x + x + x + x + x + 1 − x + 1 =
2x + 2x + 2
=
=
(x − 1)(2 x + 2 x + 2)
(x − 1)(2 x + 2 x + 2) x − 1
2(x − 1)(x + x + 1)
(
3
18 x − 1
2
3
3
4x2 + 4x + 4
3
2
h. Evaluate the integral
2
2
3
2
3
2
3
2
2
3
3
2
3
1
∫ x4 − 1 dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is an improper rational
fraction use synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational
fraction.
(
)(
)
(
)
Second - Factor the denominator x 4 − 1 into x 2 − 1 x 2 + 1 = (x − 1) (x + 1) x 2 + 1 .
Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the
integrand can be represented in the following way:
1
x4 − 1
=
1
=
(x − 1) (x + 1) (x + 1)
2
A
B
Cx + D
+
+
x − 1 x + 1 x2 + 1
Fourth - Solve for the constants A , B , C and D by equating coefficients of the like powers.
1
4
x −1
=
(
)
(
)
A(x + 1) x 2 + 1 + B (x − 1) x 2 + 1 + (x − 1) (x + 1) (Cx + D )
(
(x − 1) (x + 1) (x 2 + 1)
(
)
)
1 = A(x + 1) x 2 + 1 + B(x − 1) x 2 + 1 + (x − 1)(x + 1)(Cx + D )
1 = Ax3 + Ax 2 + Ax + A + Bx3 − Bx 2 + Bx − B + Cx3 + Dx 2 − Cx − D
1 = ( A + B + C )x3 + ( A − B + D )x 2 + ( A + B − C )x + ( A − B − D ) therefore,
A+ B+C = 0
A− B+ D = 0
A− B− D =1
A+ B−C = 0
1
1
1
which result in having A = , B = − , C = 0 , and D = −
4
2
4
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
A
B
Cx + D
1
1
1
1
1
1
∫ x4 − 1 dx = ∫ x − 1 dx + ∫ x + 1 dx + ∫ x2 + 1 dx = 4 ∫ x − 1 dx − 4 ∫ x + 1 dx − 2 ∫ x2 + 1 dx
Sixth - Integrate each integral individually using integration methods learned in previous sections.
1
1
1
1
1
1
1
1
1
dx −
dx −
dx = ln x − 1 − ln x + 1 − tan −1 x + c
4 x −1
4 x +1
2 x2 + 1
4
4
2
∫
∫
∫
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
=
1 1
1 1
1
1
1
1
1
⋅1 − ⋅
⋅1 − ⋅ 2
⋅1 + 0
ln x − 1 − ln x + 1 − tan −1 x + c , then y ′ = ⋅
4
4
2
4 x −1
4 x +1
2 x +1
1
1
1
−
−
4(x − 1) 4(x + 1) 2 x 2 + 1
(
2
− 2x − 2x + 2x + 2
(
2
)(
4 x −1 4 + x
2
)
i. Evaluate the integral
=
=
)
(
4
4
)
4 x −1
=
(x + 1) (x 2 + 1) − (x − 1) (x 2 + 1) − 2(x − 1) (x + 1) = x3 + x + x 2 + 1 − x3 − x + x 2 + 1
4(x 2 − 1)(4 + x 2 )
4(x − 1) (x + 1) (x 2 + 1)
1
x4 − 1
1
∫ x3 + 64 dx .
First - Check to see if the integrand is a proper or an improper rational fraction. If the integrand is a rational fraction use
synthetic division (long division) to reduce the rational fraction to the sum of a polynomial and a proper rational fraction.
(
)
Second - Factor the denominator x3 + 64 into (x + 4 ) x 2 − 4 x + 16 .
Hamilton Education Guides
111
Advanced Integration
Solutions
Third - Write the factors in partial fraction form. Since one of the factors in the denominator is in quadratic form, the
integrand can be represented in the following way:
1
x3 + 64
1
=
=
(x + 4) (x − 4 x + 16)
2
A
Bx + C
+
x + 4 x 2 − 4 x + 16
Fourth - Solve for the constants A , B , and C by equating coefficients of the like powers.
3
1
x + 64
(
(
)
A x 2 − 4 x + 16 + (Bx + C ) (x + 4 )
=
)
(x + 4) (x 2 − 4 x + 16)
1 = A x 2 − 4 x + 16 + (Bx + C )(x + 4 ) = Ax 2 − 4 Ax + 16 A + Bx 2 + 4 Bx + Cx + 4C
1 = ( A + B )x 2 + (− 4 A + 4 B + C )x + (16 A + 4C ) therefore,
−4 A + 4 B + C = 0
A+ B = 0
which result in having A =
16 A + 4C = 1
1
1
1
, B=−
, and C =
48
48
6
Fifth - Rewrite the integral in its equivalent partial fraction form by substituting the constants
with their specific values.
1
∫ x3 + 64
dx =
∫
A
Bx + C
1
1
dx + 2
dx =
dx +
x+4
48 x + 4
x − 4 x + 16
∫
∫
1 x+ 1
− 48
6
1
1
1
x −8
∫ x2 − 4 x + 16 dx = 48 ∫ x + 4 dx − 48 ∫ x2 − 4 x + 16 dx
Sixth - Integrate each integral individually using integration methods learned in previous
du
du
. Also, x − 8 can be
= 2 x − 4 and dx =
2x − 4
dx
sections. To solve the second integral let u = x 2 − 4 x + 16 , then
rewritten as x − 8 = (x − 2 ) − 6 =
1
(2 x − 4) − 6 . Therefore,
2
(x − 2) − 6 dx = 1 1 dx − 1
1
1
1
1
1
1
x −8
dx −
dx −
dx =
48 x + 4
48 x + 4
48 x 2 − 4 x + 16
48 x 2 − 4 x + 16
48 x + 4
48
∫
∫
∫
∫
(
)
∫ x − 4 x + 16
=
1
1
1
dx −
48 x + 4
96
=
1
48
=
1
1
12
x−2
1
1
12
x−2
ln x + 4 − ln x 2 − 4 x + 16 +
tan −1
+c =
ln x + 4 −
ln x 2 − 4 x + 16 −
tan −1
+c
48
96
8 ⋅ 12
48
96
96
12
12
∫
1
1
2x − 4
∫
1 2x − 4 − 6
2
dx
2
1
6
1
1
1
2x − 4
∫ x2 − 4 x + 16 dx + 48 ∫ x2 − 4 x + 16 dx = 48 ∫ x + 4 dx − 96 ∫ u
1
1
1
1
1
1
∫ x + 4 dx − 96 ∫ u du + 8 ∫ (x − 2)2 + 12 dx = 48 ln x + 4 − 96 ln u + 8 12 tan
⋅
du
6
+
2 x − 4 48
−1 x − 2
12
1
∫ (x − 2)2 + 12 dx
+c
Seventh - Check the answer by differentiating the solution. The result should match the integrand.
Let y =
y′ =
x−2
1
12
1
+ c , then
tan −1
ln x + 4 − ln x 2 − 4 x + 16 +
96
48
96
12
1
1
1
1
12
⋅
−
⋅
⋅ (2 x − 4 ) +
⋅
48 x + 4 96 x 2 − 4 x + 16
96
1
(1 ⋅ 12 )− 0 ⋅ (x − 2) + 0 = 1 − x − 2
48(x + 4 ) 48(x − 4 x + 16 )

( 12 )

1 +  x − 2
 12 
2
⋅
2
+
x−2
1
6
12
12
12
=
−
+
⋅
⋅
2
48(x + 4 ) 48 x 2 − 4 x + 16
96 12 + (x − 2 )2 12
48 x − 4 x + 16
=
1
−x + 8
+
48(x + 4 ) 48 x 2 − 4 x + 16
=
(
(
16 + 32
(
48(x + 4 ) x 2 − 4 x + 16
)
Hamilton Education Guides
)
=
=
)
(
2
=
)
−x + 2 + 6
1
+
48(x + 4 ) 48 x 2 + 4 x + 16
(
(x − 4 x + 16)+ (− x + 8)(x + 4) = x − 4 x + 16 − x − 4 x + 8x + 32
48(x + 4 ) (x − 4 x + 16 )
48(x + 4 ) (x − 4 x + 16 )
2
2
2
2
(
48
48(x + 4 ) x 2 − 4 x + 16
)
2
)
=
3
2
1
2
x − 4 x + 16 x + 4 x − 16 x + 64
=
3
1
x + 64
112
Advanced Integration
Solutions
Section 1.4 Practice Problems – Integration of Hyperbolic Functions
1. Evaluate the following integrals:
a. Given
du
d
du
∫ cosh 3x dx let u = 3x , then dx = dx 3x = 3 which implies dx = 3 . Therefore,
du
1
1
1
∫ cosh 3x dx = ∫ cosh u ⋅ 3 = 3 ∫ cosh u du = 3 sinh u + c = 3 sinh 3 x + c
Check: Let y =
b. Given
1
d
1
1
1 d
d
sinh 3 x + c , then y ′ = ⋅ sinh 3 x +
c = ⋅ cosh 3 x ⋅
3 x + 0 = ⋅ cosh 3 x ⋅ 3 = cosh 3 x
3
3
dx
3
3 dx
dx
∫ (sinh 2 x − e ) dx = ∫ sinh 2 x dx + ∫ e dx let:
3x
3x
a. u = 2 x , then
du
du
d
du
and
= 2 ; du = 2dx ; dx =
=
2x ;
dx
dx dx
2
b. v = 3 x , then
dv
dv
d
dv
.
=
3x ;
= 3 ; dv = 3dx ; dx =
dx dx
3
dx
=
du
1
dv
1
1
1 v
∫ sinh 2 x dx + ∫ e dx = ∫ sinh u ⋅ 2 + ∫ e ⋅ 3 = 2 ∫ sinh u du + 3 ∫ e dv = 2 cosh u + c1 + 3 e + c2
3x
Therefore,
3x
v
1
1
1
1
cosh 2 x + e3 x + c1 + c2 = cosh 2 x + e 3 x + c
3
2
3
2
Check: Let y =
=
c. Given
1
1
1 d
1 d
d
1
1
d
d
cosh 2 x + e3 x + c then y ′ = ⋅ cosh 2 x + ⋅ e3 x +
c = ⋅ sinh 2 x ⋅
2 x + ⋅ e3 x ⋅ 3 x + 0
2
3
2 dx
3 dx
dx
3
2
dx
dx
1
1
2
3
⋅ sinh 2 x ⋅ 2 + ⋅ e3 x ⋅ 3 = ⋅ sinh 2 x + ⋅ e3 x = sinh 2 x + e3 x
2
3
2
3
du
d
du
∫ csc h 5x dx let u = 5x , then dx = dx 5x = 5 which implies du = 5dx ; dx = 5 . Therefore,
du
1
1
1
u
5x
∫ csc h 5x dx = ∫ csc h u ⋅ 5 = 5 ∫ csc h u du = 5 ln tanh 2 + c = 5 ln tanh 2 + c
Check: Let y =
1
1
d 
5x 
1
5x
d
5x
1 d
⋅  tanh
ln tanh
+ c , then y ′ = ⋅ ln tanh
c = ⋅
+
+0
5
2
dx
2
5 dx
5 tanh 5 x dx 
2 
2
2 5x
1 5 sec h 2
1
1

2 5x 5 
= ⋅
=
⋅
⋅
sec
⋅
+
0
h


10 tanh 5 x
5 tanh 5 x 
2 2
2
2
2 5x
1 sec h 2
⋅
2 tanh 5 x
2
=
=
2 5x
1 1 − tanh 2
⋅
2
tanh 52x
1
= ⋅
2
1−
sinh 2 5 x
2
cosh 2 5 x
sinh 5 x
2
2
cosh 5 x
2
cosh 2 5 x − sinh 2 5 x
2
2
cosh 2 5 x
2
sinh 5 x
2
2⋅
cosh 5 x
2
=
d. Given
=
1
cosh 2 5 x
=
2
sinh 5 x
2
2⋅
cosh 5 x
2
5x
cosh 2
1
1
1
1
=
=
=
= csc h5 x
⋅
5
x
5
x
5
x
sinh
5x
2 cosh 2 5 x ⋅ sinh 5 x
2 cosh 2 ⋅ sinh 2
sinh 2 ⋅ 2
2
2
du
d
du
du
∫ x sec h x dx let u = x , then dx = dx x ; dx = 3x ; du = 3x dx ; dx = 3x2 . Therefore,
2
3
2 3
du
1
3
2
2
1
1
∫ x sec h x dx = ∫ x sec h u ⋅ 3x2 = 3 ∫ sec h u du = 3 tanh u + c = 3 tanh x + c
2
Check: Let y =
e. Given
2
2 3
2
2
3
1
1
d 3
1
1 d
d
tanh x3 + c , then y ′ = ⋅ tanh x3 +
c = ⋅ sec h 2 x3 ⋅
x + 0 = ⋅ sec h 2 x3 ⋅ 3 x 2 = x 2 sec h 2 x3
3
3 dx
dx
3
3
dx
∫ 3 x csc h (x + 1)dx let u = x + 1 , then dx = dx (x + 1); dx = 4x ; du = 4 x dx ; dx = 4x3 . Therefore,
2
3
2
4
Hamilton Education Guides
4
du
d
4
du
3
3
du
113
Advanced Integration
Solutions
∫ 3 x csc h (x + 1)dx = 3 ∫ x csc h u ⋅ 4 x3 = 6 ∫ csc h u du = − 6 coth u + c = − 6 coth (x + 1) + c
2
3
2
2
4
3
(
du
2
1
1
1
2
)
(
)
4
(
) (
)
1
1 d
d
1
d 4
Check: Let y = − coth x 4 + 1 + c , then y ′ = − ⋅ coth x 4 + 1 + c = − ⋅ − csc h 2 x 4 + 1 ⋅
x +1 + 0
6 dx
dx
6
dx
6
(
f. Given
∫
(
)
)
(
)
4 x3
1
2
⋅ csc h 2 x 4 + 1 = x3 csc h 2 x 4 + 1
⋅ csc h 2 x 4 + 1 ⋅ 4 x3 =
6
3
6
=
∫ x csc h (2 x + 5)dx let u = 2 x + 5 , then dx = dx (2 x + 5); dx = (2 x + 5) ; du = 8x dx ; dx = 8x3 . Therefore,
(
du
4
4
3
)
x3 csc h 2 x 4 + 5 dx =
∫
x3 csc h u ⋅
du
8x
=
3
1
8
∫
d
du
4
csc h u du =
du
3
4
2 x4 + 5
1
1
u
+c
ln tanh
+ c = ln tanh
2
8
8
2
d
1
1
2x4 + 5
1
1 d
2x4 + 5
d
2x4 + 5
⋅ tanh
+0
ln tanh
+ c = ⋅
+ c , then y ′ = ⋅ ln tanh
4
8 tanh 2 x + 5 dx
2
8
8 dx
2
dx
2
Check: Let y =
2
4
4
4
2 2 x +5
4
4
sec h 2 2 x 2 + 5 8 x3
x3 1 − tanh
x3
2 2x + 5 d 2x + 5
2
⋅
⋅
+
h
sec
0
⋅
⋅
⋅
=
=
=
=
4
4
4
2
dx
2
2
2
8 tanh 2 x + 5 2
8 tanh 2 x + 5
tanh 2 x + 5
1
2
2
1−
2
sinh 2 2 x + 5
2
4
cosh 2 2 x + 5
4
2
sinh 2 x + 5
2
4
cosh 2 x + 5
2
x3
⋅
=
2
4
4
cosh 2 2 x + 5 − sinh 2 2 x + 5
2
2
2 2 x 4 +5
1
2 2 x 4 +5
2
2 x 4 +5
x3 cosh 2
=
⋅
2 sinh 2 x 4 + 5
cosh 2 x + 5
cosh 2 x + 5
cosh
sinh
2
4
2
4
2
=
x
3
4
sinh 2 ⋅ 2 x 2 + 5
=
4
cosh 2 x 2 + 5
x3
x3
⋅
=
=
4
4
2 cosh 2 2 x 4 + 5 ⋅ sinh 2 x 4 + 5
2 cosh 2 x 2 + 5 ⋅ sinh 2 x 2 + 5
2
2
2
(
x
(
3
4
sinh 2 x + 5
= x3 csc h 2 x 4 + 5
)
)
du
d
du
du
∫ cosh ( x + 1) sinh ( x + 1)dx let u = cosh (x + 1) , then dx = dx cosh (x + 1) ; dx = sinh (x + 1) ; dx = sinh (x + 1) . Thus,
7
g. Given
du
1
1
∫ cosh ( x + 1) sinh ( x + 1)dx = ∫ u sinh (x + 1) ⋅ sinh (x + 1) = ∫ u du = 8 u + c = 8 cosh (x + 1) + c
7
7
Check: Let y =
h. Given
7
8
8
1
1
cosh8 (x + 1) + c , then y ′ = ⋅ 8 cosh 7 (x + 1) ⋅ sinh (x + 1) + 0 = cosh 7 (x + 1) sinh (x + 1)
8
8
du
d
du
du
∫ csc h (5x + 3) coth (5x + 3) dx let u = 5x + 3 , then dx = dx (5x + 3) ; dx = 5 ; du = 5dx ; dx = 5 . Therefore,
du
1
1
1
∫ csc h (5x + 3) coth (5x + 3) dx = ∫ csc h u coth u 5 = 5 ∫ csc h u coth u du = − 5 csc h u + c = − 5 csc h (5 x + 3) + c
1
d
1
csc h (5 x + 3) coth (5 x + 3)
Check: Let y = − csc h (5 x + 3) + c , then y ′ = − ⋅ − csc h (5 x + 3) coth (5 x + 3) ⋅ (5 x + 3) + 0 =
⋅5
5
5
5
dx
i. Given
∫e
=
5
csc h (5 x + 3) coth (5 x + 3) = csc h (5 x + 3) coth (5 x + 3)
5
∫e
x +1 sec h e x +1 dx let u = e x +1 , then du = d e x +1 ; du = e x +1 ; du = e x +1 ⋅ dx ; dx = du . Therefore,
dx dx
dx
e x +1
x +1 sec h e x +1 dx =
(
∫e
x +1
sec h u ⋅
)
du
e
x +1
=
Check: Let y = sin −1 tanh e x +1 + c , then y ′ =
Hamilton Education Guides
−1
−1
x +1
∫ sec h u du = sin (tanh u ) + c = sin ( tanh e ) + c
1
1 − tanh 2 e x +1
⋅
d
tanh e x +1 + 0 =
dx
sec h 2e x +1
sec h 2e x +1
⋅
d x +1
e
dx
114
Advanced Integration
Solutions
sec h 2e x +1
=
sec h e x +1
⋅ e x +1 = e x +1 sec h e x +1
2. Evaluate the following integrals:
a.
d
du
du
du
∫ tanh x sec h x dx let u = tanh x , then dx = dx tanh x ; dx = sec h x ; du = sec h x dx ; dx = sec h2 x . Thus,
5
2
du
1
∫ tanh x sec h x dx = ∫ u ⋅ sec h x ⋅ sec h2 x = ∫ u du = 5 + 1 u
5
Check: Let y =
b. Given
5
2
2
5
2
2
5 +1
+c =
1
1 6
u + c = tanh 6 x + c
6
6
1
1
tanh 6 x + c then y ′ = ⋅ 6 (tanh x ) 6 −1 ⋅ sec h 2 x + 0 = (tanh x ) 5 sec h 2 x = tanh 5 x sec h 2 x
6
6
d
du
du
∫ coth (x + 1) csc h ( x + 1) dx let u = coth (x + 1) , then dx = dx coth (x + 1) ; dx = − csc h (x + 1) c
6
2
; du = − csc h 2 (x + 1) dx ; dx = −
du
csc h 2 (x + 1)
. Thus,
2
− du
∫ coth (x + 1) csc h ( x + 1) dx = ∫ u ⋅ csc h (x + 1) ⋅ csc h2 (x + 1)
6
2
6
2
1
1
= − u 6 du = − u 7 + c = − coth7 ( x + 1) + c
7
7
∫
1
1
Check: Let y = − coth 7 (x + 1) + c then y ′ = − ⋅ 7[ coth (x + 1) ] 7 −1 ⋅ − csc h 2 (x + 1) + 0 = coth 6 (x + 1) csc h 2 (x + 1)
7
7
c. Given
3x
du
du
1
d
∫ e tanh e dx let u = e , then dx = dx e
3x
3x
3x
;
du
du
= 3e3 x ; du = 3e3 x dx ; dx = 3 x . Therefore,
dx
3e
1
1
∫ e tanh e dx = ∫ e tanh u ⋅ 3e3x = 3 ∫ tanh u du = 3 ln cosh u + c = 3 ln cosh e
3x
Check: Let y =
d. Given
3x
3x
3x
+c
3
1
1
1
ln cosh e3 x + c , then y ′ = ⋅
⋅ sinh e3 x ⋅ 3e3 x + 0 = e3 x tanh e3 x = e3 x tanh e3 x
3
3
3 cosh e3 x
∫ x sec h ( x + 1) dx let u = (x + 1) , then dx = dx (x + 1) ; dx = 4x ; du = 4 x dx ; dx = 4x3 . Thus,
3
4
du
4
d
du
4
du
3
3
∫ x sec h ( x + 1) dx = ∫ x ⋅ sec h u ⋅ 4 x3 = 4 ∫ sec h u ⋅ du = 4 sin (tanh u ) + c = 4 sin [tanh (x + 1) ]+ c
3
Check: Let y =
×
e. Given
du
3
4
[ (
1
1
)]
1 −1
sin tanh x 4 + 1 + c then y ′ =
4
(
(
1
−1
1
(
)
⋅
4 1 − tanh 2 x 4 + 1
)
)
(
−1
4
)
d
tanh x 4 + 1 + 0 =
dx
( )
4 sec h (x + 1)
sec h 2 x 4 + 1
2
4
sec h 2 x 4 + 1
4 x3
d 4
⋅ 4 x3 =
⋅ sec h x 4 + 1 = x3 sec h x 4 + 1
x +1 =
4
4
dx
4 sec h x + 1
(
)
(
du
(
)
d
)
du
du
∫ sec h ( 3x + 2) dx let u = 3x + 2 , then dx = dx ( 3x + 2) ; dx = 3 ; du = 3dx ; dx = 3 . Thus,
du
1
1
1
∫ sec h ( 3x + 2) dx = ∫ sec h u ⋅ 3 = 3 ∫ sec h u du = 3 sin (tanh u ) + c = 3 sin [ tanh (3 x + 2) ] + c
Check: Let y =
×
f. Given
1 −1
sin [ tanh (3 x + 2 ) ] + c , then y ′ =
3
−1
1
3 1 − tanh 2 (3 x + 2 )
⋅
−1
d
tanh (3 x + 2 ) + 0 =
dx
sec h 2 (3 x + 2 )
3 sec h 2 (3 x + 2 )
2
2
d
(3x + 2) = sec h (3x + 2) ⋅ 3 = 3 ⋅ sec h (3x + 2) = sec h (3x + 2)
dx
3 sec h (3 x + 2 )
3 sec h (3 x + 2 )
∫e
cosh (3 x + 5 )
sinh (3 x + 5) dx let u = cosh (3 x + 5) , then
Hamilton Education Guides
du
d
du
d
= sinh (3 x + 5) ⋅ (3 x + 5)
=
cosh (3 x + 5) ;
dx dx
dx
dx
115
Advanced Integration
;
Solutions
du
du
. Therefore,
= sinh (3 x + 5) ⋅ 3 ; dx =
dx
3 sinh (3 x + 5)
∫e
cosh (3 x + 5 )
sinh (3 x + 5) dx =
Check: Let y =
du
1
1
1 u
∫ e sinh (3x + 5) ⋅ 3sinh (3x + 5) = 3 ∫ e du = 3 e + c = 3 e
u
u
cosh (3 x +5 )
+c
d
1
3
1 cosh (3 x +5 )
e
+ c , then y ′ = ⋅ ecosh (3 x + 5 ) ⋅ sinh (3 x + 5) ⋅ (3 x + 5) + 0 = ⋅ ecosh (3 x + 5 ) ⋅ sinh (3 x + 5)
dx
3
3
3
= ecosh (3 x + 5 ) sinh (3 x + 5)
g.
∫ tanh x dx = ∫ tanh x tanh x dx = ∫ tanh x (1 − sec h x )dx = − ∫ tanh x sec h x dx + ∫ tanh x dx . To solve the first
3
5
2
3
3
2
3
du
d
du
du
. Therefore,
= sec h 2 x ; du = sec h 2 x dx ; dx =
=
tanh x ;
dx
dx dx
sec h 2 x
integral let u = tanh x , then
∫
∫
2
− tanh 3 x sec h 2 x dx = − u 3 sec h 2 x ⋅
du
sec h 2 x
1
1
1
= − u 3du = − u 4 + c = − tanh 4 x + c . In Example 5.4-6, problem
4
4
∫
∫ tanh x dx = − 2 tanh x + ln cosh x + c . Therefore,
3
letter d, we found that
2
∫ tanh x dx = ∫ tanh x tanh x dx = ∫ tanh x (1 − sec h x ) dx = − ∫ tanh x sec h x dx + ∫ tanh x dx
5
= −
3
2
3
3
2
2
3
1
1
1
1

 1
tanh 4 x + tanh 3 x dx = − tanh 4 x +  − tanh 2 x + ln cosh x + c  = − tanh 4 x − tanh 2 x + ln cosh x + c
4
4
2
4

 2
∫
4
2
sinh x
1
1
Check: Let y = − tanh 4 x − tanh 2 x + ln cosh x + c , then y ′ = − tanh 3 x ⋅ sec h 2 x − tanh x ⋅ sec h 2 x +
+0
4
2
4
2
cosh x
(
(
)
)(
= − tanh 3 x sec h 2 x − tanh x sec h 2 x + tanh x = − sec h 2 x tanh 3 x + tanh x + tanh x = − 1 − tanh 2 x tanh 3 x + tanh x
(
)
)
+ tanh x = − tanh 3 x + tanh x − tanh 5 x − tanh 3 + tanh x = − tanh 3 x − tanh x + tanh 5 x + tanh 3 + tanh x = tanh 5 x
h.
∫ coth x dx = ∫ coth x coth x dx = ∫ coth x (1 + csc h x ) dx = ∫ coth x csc h x dx + ∫ coth x dx . To solve the first
3
5
2
2
3
2
2
3
du
d
du
du
. Thus,
= − csc h 2 x ; du = − csc h 2 x dx ; dx = −
=
coth x ;
dx
dx dx
csc h 2 x
integral let u = coth x , then
3
3
du
1
1
∫ coth x csc h x dx = ∫ u csc h x ⋅ − csc h 2 x = − ∫ u du = − 4 u + c = − 4 coth x + c . In example 5.4-6, problem
3
2
1
3
4
4
∫ coth x dx = − 2 coth x + ln sinh x + c . Grouping the terms together we have
3
letter g, we found that
2
∫ coth x dx = ∫ coth x coth x dx = ∫ coth x (1 + csc h x )dx = ∫ coth x csc h x dx + ∫ coth x dx = − 4 coth x + ∫ coth x dx
5
= −
3
2
3
3
2
2
1
3
4
3
1
1
1
 1

coth 4 x +  − cot 2 x + ln sinh x + c  = − coth4 x − coth 2 x + ln sinh x + c
4
2
4
2


− 4 ⋅ coth 3 x ⋅ − csc h 2 x 2 ⋅ coth x ⋅ − csc h 2 x cosh x
1
1
Check: Let y = − coth 4 x − coth 2 x + ln sinh x + c , then y ′ =
−
+
+0
4
2
sinh x
4
2
(
(
)
)(
)
= coth 3 x csc h 2 x + coth x csc h 2 x + coth x = csc h 2 x coth 3 x + coth x + coth x = coth 2 x − 1 coth 3 x + coth x + coth x
= coth 5 x + coth 3 x − coth 3 x − coth x + coth x = coth 5 x
i.
∫ coth x dx = ∫ coth x coth x dx = ∫ coth x (1 + csc h x ) dx = ∫ ( coth x csc h x + coth x ) dx = ∫ coth x csc h x dx .
6
4
∫
2
4
2
+ coth x dx . In example 5.4-6, problem letter e, we found that
Hamilton Education Guides
4
2
1
4
4
2
∫ coth x dx = − 3 coth x − coth x + x + c . Therefore,
4
3
116
Advanced Integration
Solutions
∫ coth x dx = ∫ coth x coth x dx = ∫ coth x (1 + csc h x ) dx = ∫ coth x csc h x dx + ∫ coth x dx = ∫ coth x csc h x dx
4
6
4
2
2
4
2
4
4
2
1
1
 1

+  − coth 3 x − coth x + x + c  = − coth5 x − coth 3 x − coth x + x + c
3
3
5


5 ⋅ coth 4 x ⋅ − csc h 2 x 3 ⋅ coth 2 x ⋅ − csc h 2 x
1
1
Check: Let y = − coth 5 x − coth 3 x − coth x + x + c , then y ′ = −
−
5
3
3
5
(
)
+ csc h 2 x + 1 + 0 = coth 4 x ⋅ csc h 2 x + coth 2 x ⋅ csc h 2 x + csc h 2 x + 1 = csc h 2 x coth 4 x + coth 2 x + 1 + 1
(
)(
)
= coth 2 x − 1 coth 4 x + coth 2 x + 1 + 1 = coth 6 x + coth 4 x + coth 2 x − coth 4 x − coth 2 x − 1 + 1 = coth 6 x
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About the Author
Dan Hamilton received his B.S. degree in Electrical Engineering from Oklahoma State
University and Master's degree, also in Electrical Engineering from the University of Texas at
Austin. He has taught a number of math and engineering courses as a visiting lecturer at the
University of Oklahoma, Department of Mathematics, and as a faculty member at Rose State
College, Department of Engineering Technology, at Midwest City, Oklahoma. He is currently
working in the field of aerospace technology and has published several books and numerous
technical papers.
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